All Triangles are Equilateral - Numberphile

[@calvinscheuerman]

I love troll math like this. it reminds me of the pi = 4 thing that was going around the internet a few years ago. it's just fun. wrong, but fun.

[@forlorneater6595]

I call it monkey math

[@mgsquared5204]

@@doronyaniv9205 you start with a square with side length 1 and bring the corners in so they touch a circle with diameter 1. Then you take the new corners generated and bring those in to touch the circle. You do this forever. What you get is a shape that looks exactly like a circle. However, at no point in any of the steps did the perimeter change. It is always 4. This means the circumference of a circle in relation to its diameter is 4. The thing here is you never *actually* get a circle by bringing the corners in to touch a circle. It’s just something that looks like a circle. With infinite “zooming” you’d still see the jagged corners or something like that.

[@d4slaimless]

@@mgsquared5204 Yeah, it looks fun.

[@neilfairbairn440]

Can't pi be 4 if it is a square and it approaches 3.14 as the sides approach infinity?

[@chriswebster24]

⁠​⁠​⁠​⁠@@neilfairbairn440No, that’s ridiculous.

[@roquegabrielroque]

In fact, all the following propositions are corrects: XB = XC XB* = XC* BB* = CC* XBB* = XCC* AXB* = AXC* AB* = AC* I did it again in a drawing software and I realized that if the line C of the triangle is larger than B, then the point C* is located between the points C and A, and not after the point C, like illustrated. Therefore, in this case the following equations will be: AB* - BB* = AB AC* + CC* = AC Because that, the conclusion "AB=AC" is wrong. The most that we can affirm is: AC = AB + BB* * I'm brazilian and sorry by the poor english.

[@pokegan32155]

That was fantastic. Greetings from Brazil.

[@Gee800220]

Actually AXB* = AXC* is NOT true and it is VERY easy to prove by simply going trough the rules of mirroring an object. If it would be true, than when mirroring AXB* the mirror of B should be C. That concludes from BB* = CC*. But if C is the mirror of B than the lines BC and AX should have a right angle between them, since AX is the line used for the mirroring. And a simple glance of the drawing proves that it is NOT the case.

[@roquegabrielroque]

Gee The drawing representation is wrong, so in THIS representation AXB* is not equal to AXC*. But if you follow the instructions given by the video you will get the scenario described by me. And, in that case, AXB* = AXC* is TRUE.

[@raphaeldayan]

não entendi não irmão explica ae pros BR

[@roquegabrielroque]

raphael dayan pra galera que fala português então... De fato, as seguintes expressões estão corretas: XB = XC XB* = XC* BB* = CC* XBB* = XCC* AXB* = AXC* AB* = AC* Eu refiz o exercício em um software 2D de desenho e percebi que: se a linha C do triângulo for maior que a B, então o ponto C* estará localizado entre os pontos C e A, e não depois do ponto C, como foi ilustrado no vídeo. Sendo assim, as expressões corretas seriam: AB* - BB* = AB AC* + CC* = AC Por causa disso, a conclusão: "AB=AC" está errada. O máximo que podemos afirmar é: AC = AB + BB*

[@sethmorse3139]

this is learning... an educated man asking amateurs to solve something... I'd love to have my high school classes structured like this

[@EP1CStar]

I FIGURED OUT WHERE YOU WENT WRONG!! ..it's not equilateral

[@Jivvi]

Yep, that's exactly it. 😉

[@hps362]

This comment has just made my day.

[@Jivvi]

***** pretty sure it's in the second step. x²+1²/x² isn't (x+1/x)²

[@valeriobertoncello1809]

+Jivan Scarano Wow you guys are overthinking.. it's just that XBB* is NOT congruent to XCC* (they are indeed in his drawing, that is wrong on purpouse: if you look closely at 3:02 the angle that's supposed to be right inside XCC* is really NOT right)

[@plasmacrab_7473]

4x4x6 Fishercube?

[@raichakrabarti8085]

Both the points B* and C* cannot lie outside the triangle. The statement AB*=AC* is true, and BB*=CC is true, but for doing AC*-CC* and AB*-BB*, you need to have both the points outside the triangle, which has not been followed here.

[@raichakrabarti8085]

Also, the triangle that has been drawn here is almost isosceles, in which case it can be, bit had the process been followed for a triangle that is most definitely scalene, the lines would not have intersected outside the triangle.

[@SpiderWaffle]

+Rai Chakrabarti (Itrabarkach Iar) Yep, if you just check with a straight edge segment it's easy to eyeball that BM is longer than BC, (line bisector is false) and I suspect the angle bisector has BAX smaller than XAC. This is to help create the illusion that the intersection, X, MUCH further away from BC than it really is, and thus make it easy to draw the two AX projections outside of AB and AC.

[@dragonsandwich]

+Rai Chakrabarti (Itrabarkach Iar) I think you're correct. Here's a proof of why B* and C* cannot both lie outside the triangle. As shown in the video, BXB* and CXC* are congruent. That means the angle(CXC*) and angle(BXB*) are equal. We then look at the shape AB*XC*, which has two 90 degree angles. Therefore the angle(B*XC*) = 180-angle(BAC). Now, since angle(CXC*) = angle(BXB*), angle(BXC) = angle(B*XC*). We now turn our attention to the shape ABXC: angle(ABX) + angle(ACX) + angle(BXC) + angle(BAC) = 360. angle(BXC) + angle(BAC) = 180, so angle(ABX) + angle(ACX) = 180. This means that either both angle(ABX) and angle(ACX) are 90 degrees or one of them is less than 90 degrees. If one of them is less than 90 degrees, then either C* or B* is inside the triangle, thus invalidating the proof. This proof works for the case where angle(ABX) = angle(ACX) = 90.

[@1cor731]

+Rai Chakrabarti (Itrabarkach Iar) Ah, that's it. The example in the video is badly drawn; X should be about 25% closer to A, so that C* would be inside the line segment AC. Both the pairs of triangles are congruent; if we call the lengths d and e, then AB*=AC*=d and BB*=CC*=e. But while the long side AC = d+e, the shorter side AB is d-e. It's a very clever puzzle.

[@vishal240993]

+Rai Chakrabarti (Itrabarkach Iar) yup thats correct, thats why one should start with not so equilateral triangle looking thing :)

[@fetchboy123]

The problem here is that the example he used here had generalized angles. I tried recreating the problem using a protractor to draw an exact 30-60-90 triangle. I bisected the 90 degree angle and created the same diagram shown above, however, I found that (corresponding) lines on my diagram, XB* and XC*, did not fall on the outside of the triangle! In fact, line XB* fell onto line AB inside the triangle and line XC* fell on line AC outside the triangle! Using the same geometric rules used in the video, I found that line AB (which included the point B*) was equal to the line AC* (which included the point C). Once this happens, we cannot simply subtract the CC* and BB* from both sides, because that would take away part of a side of the triangle ABC! Therefore, this is FALSE!!! :)

[@PC_Simo]

Great going 👍🏻!

[@Drachenbauer]

I reproduced it too by using Inkscape (with two different starting triangles), because i can use exact snapping mechanics there to make it accurate. and i got the same thing as Shreyas: Only one of the lines, that create the points B* and C* leads outside the starting triangle like shown, the other one leads into the inside of the starting-triangle (the one, that points to the longer edge of the starting-triangle). In the video it looks like the intersection of the angle bisector and the line, that starts at point M is located too far to the bottom (he made it look like an equilateral triengle appearing there, but mine are more flattened). It looks like one or both of these two center-lines are not constructed accurate enough. I noticed, while dawing these two centerlines, he sayd things like "this is about here" so it really seems like he just eyeballed it instead of measuring. And it looks like the computer-drawn version just tried to reproduce, how his handdrawn construction looks, instead of doing theese steps geometrical accurate. I noticed one more thing: If you compare the top angle of the starting triangle with the angle between the two lines of each bottom-construction-line-pair, 90° is exactly in the middle between theese two angles. And in the case, that the top angle is 90°, theese bottom angles are 90° too.

[@AlexeyKatyshev]

All the congruency is fine since triangles with right angles. Actual mistake starts at 4.05 when it says that AB* - BB* = AB AND AC* - CC* = AC because one of C* and B* is inside and the other one is outside. It can be obviously proven since ABXC is on one circle (X is the middle of the arc BC), so ABX+ACX = 180, hence one of them is bigger that 90 when other is less(they equal only if AB=AC), hence perpendiculars XC* and XB* can't both be inside or both outside,

[@igbu]

"since ABXC is on one circle", pls to elaborate? :)

[@AlexeyKatyshev]

Igor Bućo since both perpendicular at M and bisector from A cross arc BC(of the circumcircle ABC) right in the middle.

[@igbu]

Alexey Katyshev Both what perpendicular at M? Sorry, not following yet. Yes, ABC are on the circumscribed circle of ABC triangle, with its center somewhere on the XM. And yes, a circle that contains BXC clearly has X dividing BC arc in half, but I don't see the obvious proof that this circle is *the* circumcircle of ABC.

[@AlexeyKatyshev]

Igor Bućo If you take the only one possible circle around ABC, then bisector from A cross arc BC in the middle AND perpendicular line at M also cross arc BC in the middle => they cross this circle at the same point => this point is X.

[@igbu]

I believe I get what that holds. If |BX| = |XC|, then if angles BAX and XAC are equal, A must lie on the BXC circle.

[@Nimblewright1992]

I have an elegant solution, unfortunately this KZbin comment box is too narrow to contain it.

[@MertcanEkiz]

Nimble's Last Theorem

[@Atrohumter]

Don't worry, in a few thousand years mathematicians will work their asses off to find out... xD

[@mac1991seth]

You two, sirs, just made my day.

[@annalisetrite7281]

I like you

[@ZipplyZane]

Didn't you hear? They lifted the character limit. You can write as much as you want!

[@JackLe1127]

If you drew this out carefully it'll be really easy to see that if AB is shorter than AC then the point C* will be in between A and C. This means that AB + BB* = AC - CC* and not +. Tricksy

[@CytosineTV]

Drawing out the shape doesn't help with proving him wrong. You will always make a mistake with your drawing. The shape that Carlo drew follows every axiom in geometry. It is near-impossible to draw or create something that is perfect for this. So looking at a drawing at measuring it doesn't help with your cause.

[@JackLe1127]

but he assumed things from his drawing as well

[@eduardogomes4865]

No, the thing he drew did not follow every axiom of geometry, because the assumption that the B* and C* would both lie outside the triangle is wrong.

[@xenathcytrin202]

I thought that line AX looked suspiciously not like it bisected angle A... Also angle C* is clearly not a right angle. Tricksy hobbitses.

[@alejorabirog1679]

You are totally right, I just did the drawing with ruler and compass.

[@NoriMori1992]

I was with him up until 3:21. That's when I started going "No no no no NO NO NO NO NO!"

[@donach9]

The flaw is long before that

[@zoklev]

@@donach9 I think he started going "No no no no NO NO NO NO NO!" cuz he realized that the proof would lead to the 2 sides being congruent

[@mathsx5887]

Mistake is 3:31

[@tonaxysam]

@@mathsx5887 wat, why? XB* = XC* AX = AX And they have one angle in common: alpha So they must be congruent

[@hainsay]

@@tonaxysam Yes but they are drawn at the wrong place. If BB* goes away from A, then CC* goes towards A, and vice versa.

[@ruashua]

You cannot infer that AXB* ~= AXC*. Sure, B* and C* are 90deg, BB* = CC*, and XB* = XC*, but all we know is (AB* & AC*) > (BB* & CC*). Also, the A angle bisector intercepts MX once. We know MX is an angle bisector of B*XC* because BM = CM and BX =CX. Angle AXC* does not necessarily equal AXB*. AX ix not necessarily an angle bisector of B*XC* because of its one, not infinite, intersection with MX.

[@treufuss-yt]

Yes you can. AXB* and AXC* share AX, furthermore XB* = XC* and thy have both a 90° angle on the opposide of the longer side, they are congruent.

[@ruashua]

Oh crap, you are right, idk why I didn't see that. Well, my second point still stands.....though I do not think that is the correction they are looking for.

[@Rumble-Tusk]

Treufuß False. Angles AXB and AXC aren't equal so you can't claim that just because the side lengths are the same that the triangles are congruent. Just because two triangles share two side lengths doesn't mean the triangles are congruent. If AX passed through point M then this would be a legit line of reasoning but otherwise, nope. No way, Jose.

[@ZombiePestControlInc]

Joe Leeney By definition XAB and XAC are equal (angle bisector)

[@aithne99]

Joe Leeney They are right triangles, so if they share ANY two sides, and they are both legs or one is a leg and the other is a hypothenuse, the unknown side is also the same. Check with the Pythagorean theorem. Generally, if two sides and the angle opposing the LONGER side are the same, the two triangles are congruent. Geez, this is taught in elementary school.

[@Jotakumon]

I'll write A° and B° because Google comments use the asterisk to format text into bold text. The mistake is that the picture shows B and C as always being _one the inside_, i.e. the nearest points to A compared to B° and C°. However that's not the case. If B is nearer to A than B°, then C° will be nearer to A than C. Paint a picture in _GeoGebra_ if you want to verify it, but it's what will happen. So everything is correct until the last part, where we don't have AB° = AB + BB° or we don't have AC° = AC + CC°. Which way it is depends on whether AB>AC, or AC>AB (if we knew a priori which one is bigger). To prove this we have to show that if AB>AB° AC AB°. We prove it by contradiction. We assume that AC > AC°. And what we will get (as the video has shown above) is that then AB = AC, which is a contradiction. Since AC != AC° anyway, we have that AC < AC° *Case 2:* AB < AB°. We do the same thing. So we have that *AB > AB° => AC < AC°* and that *AB < AB° => AC > AC°* which is equivalent to the reciprocal of *AB > AB° => AC < AC°*, therefore *AB > AB° AC < AC°*. *PS:* Does someone know how to write asterisks in Google comments without screwing up the text formatting?

[@FrankieSmileShow]

I think the small lie that makes this work is at 3:28. Saying those two triangles are mirror images presumes AB and AC to be equal, which is more or less the thing we are trying to prove here. Its a statement that sneakily begs the question, putting a whole lot of construction lines on there that are equal and hoping you forget the more important parts...

[@DrRemiehneppo]

well the triangles are in fact indeed mirror image of each other but the problem lies in the diagram (that we are made to assume right) and the final calculus, all the rest is true

[@chuckshunk]

DrRemiehneppo In fact, the triangles are not mirror images of each other, but rotations of each other. I believe there is a technical difference.

[@chuckshunk]

DrRemiehneppo Since the triangles are not in fact mirror images of each other but rotations, then the assertion that AB* = AC* is also wrong. In fact, AB* = C*X

[@Hamenopi]

He's creating a close mirror that is equilateral.

[@forbidd3nzz]

You can use the RHS, right hypotenuse side property, to prove that AB*X is congruent to AC*X. AX is a common side. AB*X = AC*X =90° B*X = C*X Therefore, AB*X is congruent to AC*X.

[@RedsBoneStuff]

I drew the whole thing in Geogebra to get an accurate image and found out that no matter how I move the original triangle's corners around, the following is the case: Either the point B* lies on |AB| or the point C* lies on |AC|. Unlike in your sketch, they _never both lie outside the original triangle_ (it works in your sketch because you've drawn all angles a little off from what they should be :P). Since you asked us to find the incorrection in your calculations rather than in your picture, here you go: The bottom equation at 4:17 should be AC* + CC* = AC.

[@maciejkubera1536]

I found the same. Maybe inacurate drawing is the issue. Look at red construction at 4:58. You can easly notice, that angle bisector is drawn very inacurate AND that triangles BB*X and CC*X doesn't look congruent.

[@RedsBoneStuff]

Maciej Kubera Of course they had to cheat a little with the drawing to make this work ;) Our task was to find the mistake in the *calculations*, and I found it.

[@nuthying3156]

+RedsBoneStuff You, sir need to rethink this.

[@RedsBoneStuff]

Austin Rominger Please be more specific.

[@Azouzazu]

+RedsBoneStuff CLAIM: At least one of the perpendiculars XB* or XC*, from the intersection point X of the perpendicular bisector of BC and the angle bisector of A of a non-isosceles triangle, to the other sides crosses one side inside |AB| or |AC| respectively. PROOF: Let the triangle ABC such that AB < AC < BC. Let both XB* and XC* cross AB and AC outside |AB| and |AC| respectively. Then ABC is equilateral (proof shown on the video). Then AB = AC = BC. [Contadiction!]

[@NeoJinn]

i love this guy's accent

[@trimethoxy4637]

what is it?

[@Supremebubble]

3:36 - Why are the two triangles congruent? Mirror image? It doesn't even look like a mirror image and of course isn't always a mirror image. So we can't say that these two triangles are congruent. If you want to say that it is a mirror image give the axis. But if it is a mirror image (what can happen if AX and MX lie on each other) then the triangle is equilateral. What have I won?

[@DrRemiehneppo]

nothing, the triangles are in fact congruent, don't let yourself be wronged by the diagram

[@Supremebubble]

DrRemiehneppo I may have been mistaken by the diagram (I wasn't it was just a test for you^^) but the diagram itself is the mistake. Because if B lies between A and B* then C can't lie between A and C*.

[@hoodiesticks]

But they are mirror images. AX is the axis. Alternatively, you could prove their congruency by the same rules he used with the blue triangle. Both triangles have AX for a side, they both have a right angle, and B*X = C*X. If two triangles share two side lengths and a non-inscribed angle, they are congruent.

[@bugodi327]

if you passed 9th grade geometry you know they are congruent

[@Supremebubble]

chuckshunk ***** ***** I have already mentioned that my main comment isn't right and I have already given the real solution so thanks for your comments^^ Bugod i I am still in school (A-levels) but always had mark 1. So yeah, I know they are congruent.^^

[@Jamie-st6of]

I feel that before he started his explanation, he should have shouted "LEEEEEEROOOOOOOOOY NJEEEEEENKINS!!!!

[@ThichMauXanh]

I work on automated reasoning in highschool Geometry software, so I know very well where this reasoning chain breaks down. This "paradox" pinpoints exactly what is hidden "under the rug" in Geometry education in highschool: In highschool, we are given an illusion of "rigorous" geometry proof because we are following all the axioms (congruences, angle chasing, etc). But little did we know, these axioms depend on very specific topological facts that does not get proven, but only assumed "experimentally" through inspecting the diagram. So if the diagram is incorrect, the topological assumptions are wrong, and then the axiom applications will result in incorrect facts like so. What topological fact is wrongly assumed in this diagram? It is the fact that both B* and C* are outside of the triangle. In short, what you thought of Geometry in highschool is not rigorous at all, we are doing it semi-formally by mixing axiom applications and experimentally inspecting the diagram (physically interacting with the geometry embedded in the physical universe). To fix this, we have to have axioms that deal with topological statements and make them very explicit when applying the more traditional axioms.

[@lezhilo772]

That sounds fascinating! How do you apply these topological axioms in this context? Do you have to check how many points the open sets include whenever you carry out any adding/subtracting of lengths and angles?

[@adamlatosinski5475]

The picture suggests that both B* and C* points will be outside of the triangle. Only then equations written in 4:02 will be correct. The truth is that always one of those points is outside, and the other one is inside (on one of the edges).

[@uberarius]

Yes, inacurate drawing is the problem. Right answer is AB=AB*-BB* and AC=AC*+CC*.

[@TurtleW0]

The diagram is misleading. In a correct construction, C* would actually be inside AC, so the assumption that |AC| = |AC*| - |CC*| is wrong, and |AC| = |AC*| + |CC*|

[@CurtisSmale]

I see a lot of over complicated methods of point out the error so I'll pitch in: 1) The perpendicular bisector at M will extend out to point X which is where the bisector angle at A also reaches. This is ok so far. 2) Any angle drawn from the bisector AX to B* or C* must have a common angle and any angle drawn form bisector MX to B or C must have a common angle. 3) Since these lines AX and MX are NOT parallel we know that angle AXB* - MXB != AXC* - MXC. This is because AXB* = AXC* and MXB = MXC but they do not have the same central point. That is to say that *either* part of AXB* overlaps part of MXC or part of AXC* overlaps part of MXB. So in other words, AXB* (union) MXC is greater than or less than AXC* (union) MXB but only equal in the case where AB = AC. 4) The result is that the longer of AB or AC will be the side in which B* or C* is a shortening of that line and the shorter of AB or AC will be an extension. That is to say, that if AB > AC then AB* < AB and AC* > AC. This video depicts both AB* > AB and AC* > AC which is NOT the reality for any triangle that is not isosceles on the side being examined. In the case of a triangle where at least two sides are the same, proving that is impossible since MX will be a subset of AX and thus have infinite points of intersection.

[@isithardtobevegan53]

+Curtis Smale you made it too complicated.

[@SamFisk]

+Duane that only holds if AB and AC are equal length. (Ffs replying isn't working properly..)

[@duaneediger2234]

Oh, yes, I see. I shouldn't have deleted my corrected comment (I claimed in error that M by definition is on the angle A bisector) , but I did. Thanks, Sam.

[@stopthephilosophicalzombie9017]

I like your proof the best. That was what jumped out at me also.

[@andrewberthelsen2817]

IsItHardToBeVegan? No, Curtis has outlined perfectly the only correct explanation.

[@thulyblu5486]

I have drawn a (less equal) triangle and everything is correct as presented up to the final calculus conclusion (at 3:55 ). It is stated that AB* - BB* = AB . That is not really the case. My point B* is between A and B (the line AB* is shorter than AB). So the " = AB " part is not always true, I'm not sure why it's wrong in the video, though. But still as suggested, AB* - BB* = AC* - CC* but those arent equal to AB and AC. Thoughts?

[@kwinvdv]

I came to the same conclusion, however I used the free program GeoGebra which allows you to do draw these kinds of figures but to drag points afterwards, while maintaining the same constrains, such as perpendicular lines, midpoint and angle bisector. I think the trick they used in the video is to use a triangle which is close to an isosceles triangle, such that there will be a very sharp angle between the two intersecting lines, which yields point X. In this case it would be possible that playing with the angles of the bisector and the perpendicular line a little bit will be hard to notice, but will yield a big displacement of point X, which in turns allows you to place C* on the other side of C.

[@aithne99]

Kwin van der Veen People who know what they're talking about! Finally.

[@mnkyman66332]

This is exactly it. You can prove that for any triangle ABC with AB > AC, the point B* will lie between A and B.

[@thulyblu5486]

OK, cool ^^ ... but BB* and CC* look rather far from the triangle. Is this due to 'generous' margins of error?

[@cgtoche]

Yes it is so, but the proof is not so easy... Just did it... and it's hard... At least the one I found...

[@zardzewialy]

Ok, so if M is not ON the Angle Bisector of ∠BAC, that means that ∠AXB is not equal to ∠AXC, so if these two are NOT equal, than these two triangles CAN NOT have the same lenght of |AB*| and |AC*|, as |AX| remains the same for both triangles ABX and ACX, and both |C*X| and |B*X| are equal, than all angles have to sum up to 180 degrees, and it can not happen with ∠AXB different from ∠AXC.

[@Valis52]

Except they are right triangles and we can use Pythagorean to figure the third length?

[@satyreyes]

Doesn't work, I'm afraid. Ordinarily, you're right, but in this instance the triangles are right triangles. We have that their hypotenuses are congruent, and that one of their legs are congruent; the remaining side follows by the Pythagorean Theorem, and then the triangles are congruent by SSS. Fair ball. Try something else. ;)

[@headshats]

Was just going to say that, Side Angle Side is able to prove congruence, but he used Angle Side Side which is not.

[@ACIoannina]

B* and C* are right angles, so we DO know the length of the third arm with the pythagorean theorem

[@JensDoll]

satyreyes Almost ;-) Given the length of two legs and given an angle, there are infact two triangles which can be drawn: The triangle itself and the mirror triangle! AXA* and CXC* are the same triangles, but actually have a different orientation.

[@RitvikGupta]

simple triangle AXB* and AXC* are not congurent

[@shivamchauhan8381]

StatiscopediA yes

[@shivamchauhan8381]

StatiscopediA because it cannot be proven

[@JustAlex1337]

XC*=XB* AX=AX angle AB*X= angle AC*X And it's congruent

[@joris914]

To show that triangles are congruent you need 2 equal angles and 1 equal side, not 2 equal sides and 1 equal angle. I mean, it's obvious they're not congruent from the drawn example, because just from looking you can tell that AB* is not the same length as AC*.

[@shreemanagrawal5973]

They are congruent use the rhs congruency

[@BrotherAlpha]

"A proof that all triangles are equilateral - can you see a problem with it?" AB* only equals AC* if point M is on the line AX.

[@Kabitu1]

Well, you´re basically right, but within the proof it is shown that AXB* and AC*X are congruent. So where is the flaw in the arguments?

[@densi77]

i think axb* and ac*x are not congruent. they are only congruent if we assume AB = AC...

[@AissurDrol]

Yup.

[@TheLankyNo1]

Kabitu1 Essentially at 3:31 the triangles aren't the same.

[@ElwynMoir]

Agreed. It's an instance of the begging the question fallacy. (That the proof only works if M lies on AX is the same as saying the proof only works if the original triangle is equilateral).

[@VincentEngler]

At 4:06, the video states that AB* - BB* = AB and AC* - CC* = AC... But is this true? What if AB > AB* and AC < AC*? The drawing may be throwing us off here... so lets try with some numbers... Lets Say that AB = 4, AC = 7 and BC = 10. That means that BM = CM = 5 According to the cosine rule, a² = b² + c² - 2bc cosA, we know that 10² = 7² + 4² - 2(7*4) cosA cosA = -5/8 A = 128.7° and 4² = 7² + 10² - 2(7*10) cosC cosC = 19/20 C = 18.19° so B = 180°-(128.7°+18.19°) B = 33.11° Lets call the intersection of AX and BC point Q We can solve for the lengths of BN and CN => BQ = 3.6366 CQ = 6.36405 Since BM = CM = 5, we know that M must be between C and Q and we know that QM must be CQ - CM so QM = 1.36405 We also can calculate angle AQC = 97.4597° This means that angle XQC must be 82.5403° therefore, MX = 10.4179 AND QX = 10.5068 We can also get AQ length as 2.20387 => AX length is 12.71067 Next we want to get the lengths of BX and CX. We know that BXM and CXM are right triangles with the MX leg measuring 10.4179 and the BM/CM legs measuring 5, so that means that BX and CX must be 11.5556... This means that the We further know that AB*X and AC*X are right triangles where the hypotenuse is 12.71067 and the XAB*/XAC* angles are 64.35°, so we can use some trigonometry to find that AC* = 12.71067 * cos64.35° = 5.50210... AB* = 12.71067 * cos64.35° = 5.50210... XC* = 12.71067 * sin64.35° = 11.46 XB* = 12.71067 * sin64.35° = 11.46 This lets us find that CC* = sqrt(CX² - XC*²) = sqrt(11.5556² - 11.46²) = 1.48334... BB* = sqrt(BX² - XB*²) = sqrt(11.5556² - 11.46²) = 1.48334... So that would mean that AB* - BB* should equal AB and AC*-CC* should equal AC... right? 5.50210 - 1.48334 ~ 4 So then why would AC = 7 when AC*-CC* is about 4? The answer is that AC* is shorter than AC! So the drawing makes it look like C* should fall farther from A than C, but the reality is that C* does not need to fall farther from A than C, it just needs to be at the nearest point between X and the AC leg of the BAC angle. In the case that AC < AC* and AB < AB*, we would expect our triangle to be isosceles, and in the case that AC > AC* and AB > AB* we would also expect our triangle to be isosceles. And in the case that one of these two conditions is met, and we can also do a similar construct on the B or C vertex where the distance from the vertex to the * point is either greater or less than the other two angles in both cases, then we would expect to find that the triangle is equilateral. But if they are not either both greater or both less, then the proof falls apart on the assumption that AB* - BB* = AB and AC* - CC* = AC

[@mond256]

this may be true however the chance that AC=CC* or AB=BB* is literally 1/infinity. we are working with any number existing on the number line, whole numbers only take up close to 0% because the number line is riddled with transcendental numbers and the difference between the two closest numbers are 10^-infinity. the chance that two randomly generated numbers are the EXACTLY same is literally impossible

[@1234567power1]

there's also a simpler way. do what the video does with a 3,4,5 right triangle and you'll find that line xb*/xc* (depending on which leg is 4) crosses line bc and is thus perpendicular to line ac inside the triangle which this video doesn't account for thus debunking its claim that it proves all triangles are equidistant

[@johnandrews9803]

The problem is the positioning of either C* or the B*. If AC was longer than AB, then C* would be within the line AC, so it would be AC*C as apposed to ACC*, and vise verse if AB was longer. He's just very cleverly drawn the diagram very close to an isosceles triangle, so it doesn't look too bad even though it's wrong. If you try and do this with sides that are all very different, you'll see what I mean.

[@OmegaCraftable]

I think the problem lies on assuming that the point X always exists, because if the angle bisector and the perpendicular bisector never meet it can't work surely.

[@saber1epee0]

That's a very interesting idea, and fun to play around with. First of all, if you sketch a bunch of scenarios (or with a bit of trig) you can actually prove that X will always exist at least at one point, but that's not a fun answer. The best answer is related to what you point out, because if you draw any real isosceles triangle, it will have an INFINITE amount of X's, because by definition the line AX will overlap Perfectly with line MX!

[@OmegaCraftable]

Fencer Dave I understand X will always exist in some form, but I thought maybe it had to appear on the other side of the line BC.

[@aithne99]

OmegaCraftable Yes, the "bogus step" is the position of X. Draw it :D

[@Wout12345]

Well, they can't be parallel though (don't have a formal proof, but can be visualized easily). They can be equal, but if they are, AB = AC and the theorem applies as well. The only problem with X could possible be when X lies on the other side of M, which may be possible and may mess up the further equations. My main issue IMO is though that B* and C* don't always lie outside of the triangle (in fact, I think I saw someone above proving that there's always precisely one inside and outside), which means the equations around 4:00 don't work anymore.

[@aithne99]

Fencer Dave Exactly what Wout said. This is not an idea, it's a fact that you ADD BB* to AB but SUBSTRACT CC* from AC. Or the other way around, depending on how you label things.

[@ACIoannina]

The error is at 2:00 : "Then I know that XB* is equal to XC*" . We do NOT know that! How do we know that? All we know is that XB=XC and that it's a right triangle but we know nothing about how XB* compares to XC*

[@swingardium706]

The lines XB and XC are equal because of their relationship to the midpoint of BC, but AC* and AB* are also equal; the triangles are AXB* and AXC*, AX=AX, XB=XC therefore XB*=XC*.

[@DouglaumMG]

SbAsAlSe HONRe Why are AC* and AB* equal? They are not.

[@simplesmilerwastaken]

Nope, that is not correct. Every point on angle bisector is equidistant from both sides of an angle.

[@DouglaumMG]

Denis Karabaza Not to ALL points.

[@swingardium706]

DouglaumMG Yeah, I wasn't very clear earlier; if we know that sides XB and XC are equal by definition, and we know that angles B* and C* are equal by definition, AND we know that angles alpha are equal by definition, we know that the third set of sides/ angles must also be equal :)

[@sheepersheepsheep]

The problem here starts at 3:28 where Carlo says that AXB* is a 'mirror image' of AXC*. However, they are not congruent, because it does not follow the SAS rule. side B*X=C*X (given), side AX = AX (common), but angle B*XA does not equal C*XA. We know this because since MX is the angle bisector of angle BXC, angle BXM must be equal to angle CXM. Therefore, angle B*XM is equal to C*XM (angle BXB* is equal to angle CXC* because of matching angles of congruent triangles). Since B*XM is not the same angle as BXM, and C*XM is not the same angle as CXM, then B*XM cannot be equal to C*XM. Now, since those two angles are not equal, the two green triangles are not congruent or 'mirror images', meaning that AB* is not equal to AC*, and therefore AB is not equal to AC.

[@TheCouchpotat0]

Jeez, did anyone actually pay attention in their trigonometry class? You can't declare XBB* and ACC* congruent based on the data we have - the two sides are similar, yes, but it's the angle BETWEEN those similar sides that matters in the proof of congruency, not just any arbitrary angle. (there's a reason why the corresponding rule of congruence is called Side, Angle, Side, not Side, Side, Angle) Picture it like this: imagine that X is a nail that pins two sticks XB and XB* to the wall, and at the end of XB* there's another stick fixed at right angle. With this setup, you can rotate XB* with the protraction at the side and get an infinite number of DIFFERENT triangles that all share those same characteristics. As we can see, this is no basis to proclaim that two triangles are similar. Case dismissed. In the future just try googling basic trigonometry before arguing about the drawings

[@scoldingMime]

Yeah. He claimed that the two were congruent because of Side-Side-Angle, which is not one of the five triangle congruence proofs. I'm glad somebody picked up on that one.

[@Skyhmia]

In this case it uses RHS (Right-Hypotenuse-Side). This works because Pythagoras' Theorem can be used to find the remaining side and then it is a case of SAS :)

[@treufuss-yt]

Well. It seems you only learned three of the four Statements. The fourth one says: If two sides have equal length, and the angles at the opposite of the longer side are the same, the triangles are congruent. In this case its even more special. Because of the 90° angle you can use Pythagoras to calculate the third side. Then you can use the Side-Side-Side version.

[@stormsurge1]

Think about it dude if you know all the angles (if you know 2 you know the last one) and you have one side you can draw only one triangle and you can even calculate the sides

[@deadalnix]

Dude, you just proved pythagoras wrong. You may want to apply for a Fields medal.

[@AlanKey86]

*Suspicious Things* Pause at 4:45. That "right angle" in the top left looks mighty obtuse to me! It's clever how the initial "arbitrary triangle" was chosen. The fact it's so close to being isoceles conceals the trickery - just like that illusion where 4 coloured pieces of a "right angled triangle" are shuffled about to open up a mysterious, empty square. It works because the imperfection is imperceivable.

[@LedeEleven]

The only flaw is assuming a-b-b* and a-c-c*. Everything else is done correctly, but that means that either AB*-BB*=/=AB or AC*-CC*=/=AC, which destroys the proof.

[@randylai-yt]

As many have pointed out, the figure is misleading, C* should be in between A and C. In fact, it can be shown that ABXC is a cyclic quadrilateral, so ABX + ACX = 180. Hence, if ABX is not 90, they will be different, one will be larger than 90 and the other will be less than 90. Resulting either B* in between AB or C* in between AC.

[@rjmari]

C* does NOT have to be between A and C. C* is defined as the point on this line that is perpendicular to point X which lies on the angle bisector. Sure, if you super-exaggerate, then C* CAN be between A and C, but it doesn't have to be, as evidenced in the diagram from the video. And no, the diagram in the video is not drawn inaccurately.

[@randylai-yt]

I was referring to the specific triangle shown in the video. If you draw in scale, C* will be in between A and C.

[@rjmari]

Randy Lai There is no "scale" to draw in, as no values are assigned to the lengths of the lines. The only thing that can be corrected is the location of C*, which is a little off. Even if it was moved to the correct position to form a right angle, it would still definitely by outside of AC.

[@randylai-yt]

rjmari True. In general, there no "scale". My first sentence was to point out that the figure is misleading. And the explanations followed was made under general setup.

[@rjmari]

Randy Lai Actually, I just checked and you are correct. Scale doesn't matter in this case.

[@Zachary_Setzer]

The error is the part from 4:31 through about 5:05, where he says you can do it again "on two other sides." But if you try that with, for example, angle C, A will lie on segment CA*, while B* will lie on segment CB, meaning that the final calculation beginning at 3:49 does not apply to the new construction. As others have pointed out, that calculation only works because the triangle is, in fact, isosceles as the proof shows, with AB = BC. That fact is deliberately obscured by a flawed drawing that deceitfully shows the angle bisector and the perpendicular bisector as separate when they should overlap. But the flaw in the drawing isn't the error. It just misdirects your attention away from the actual problem and gets you to spend hours trying to find a flaw where there is none. Without the flaw in the drawing, you would see that the proof, as far as it goes, looks correct and move on to the next logical step of rotating the triangle and seeing if it holds true for the other angles. You would quickly see that it does not, and the internet would miss out on quite a lot of silly people arguing for nonexistent flaws.

[@Hugh.Manatee]

Sorry, no, that's not it. The trick happens much earlier in the "proof".

[@Zachary_Setzer]

AdenineMonkey Hello, silly person arguing for a nonexistent flaw.

[@Hugh.Manatee]

Zachary Setzer I like your style, but you're still wrong. =D The angle bisector and the perpendicular bisector DO overlap in a isosceles triangle, but in any other triangle they do not and there is a single point X.

[@Zachary_Setzer]

AdenineMonkey I'm not wrong. And you haven't presented an argument. Of course there is a single point X where the respective bisectors overlap in a non-isosceles triangle. But if the triangle is not isosceles, and more specifically, if angle A is not isosceles, you will have ABB* on one side and AC*C on the other side, or vice versa. In either case, the proof does not yield AB = AC. I just read through your other comments to this video and can't even figure out what you think the flaw actually is. Please post it clearly and concisely in one place.

[@Hugh.Manatee]

Zachary Setzer I didn't because I didn't want to give the answer away in case people were still puzzling. So Spoiler alert. The flaw is that for any triangle the angle bisector runs through the centre of the opposing line. This means that XM = 0, so there are no congruent triangles, unless it's an isosceles triangle, in which case there are infinite congruent triangles (X can be any point along the angle bisector).

[@ToastyOs]

The error is in the statements at 4:03. To make the statements AB*-BB*=AB and AC*-CC*=AC is to assume that both AB* and AC* are respectively longer than AB and AC, which is not necessarily the case. In fact, unless AB and AC are equal, one of AB* or AC* must be shorter than it's respective AB or AC. The drawing's roughness misleads you into thinking that both B* and C* are outside the original triangle.

[@Christopher-yn3sk]

I see it this way. You can always draw 2 lines of equal length, that go through some point and two points on some third line, so lines XC and XB can be equal but that doesn't mean that points C and B are at the same distance from A, you can draw another line that goes through X and some point on either of the lines that goes through the AB or AC, that is equal to the XC or XB and at the same distance form the A as either one of them. That said, if you would chose that right line, the rest of the math is correct , but you would work with negative numbers, because the intercession point of the correct line would lie on one of the triangle sides AB or AC, so AB doesn't have to be equal to AC

[@Crazy_Diamond_75]

Angle AXB* is not equal to Angle AXC*, so his second congruency--where he says Tri AXB* and Tri AXC* are mirror images--doesn't work. (He also never explains why he assumed they were congruent.) Instead, Angle MXB* is equal to Angle MXC*, since Tri BMX ~= Tri CMX (SSS) and Ang BXB* ~= CXC* (angles in congruent right triangles). If Seg MX coincided with Seg AX then Tri ABC would be isoceles (AB ~= AC). After that, he would have to show that M2Y/M3Z and BY/CZ coincide on another side to prove it equilateral.

[@jonathandean9498]

I totally agree that is the correct answer

[@GaganDeepSingh0123]

Correct answer :)

[@cristinapopisca5199]

the angles AXB*=AXC* because: 1. any tri has a total angle sum. of 180.2. If BAX= CAX= a => B*AX=C*AX. 3. XB*A=XC*A= 90. If 1, 2, 3 true=> AXC* = 180- a - 90 and AXB* = 180 - a - 90=> AXC*=AXB*

[@EagleMercer69]

Cristina Popisca equal angles does not prove congruency though, it only proves similarity, they're not the same thing

[@Crazy_Diamond_75]

Cristina Popisca You're right, my mistake. It's Ang AXB not being equal to Ang AXC that is the issue. The main problem I was trying to point out I guess is that while Segment AX bisects Ang BAC, it does not bisect Ang BXC, but now I'm trying to figure out where the problem really is lol.

[@kyleburt1581]

I’m pretty sure it’s because for a triangle to be equilateral the angles have to be the same but they aren’t the same if your angle bisector and perpendicular bisector don’t line up completely. If the intersect outside the triangle then they not all of the angles are correct and the triangle cannot possibly be equilateral

[@fahrenheit2101]

He wasnt asking that. He was asking you to find the flaw in his proof.

[@jyxtheberzerking4824]

I got it: triangle AB*X can't be congruent to triangle AC*X because it doesn't reflect across bisector M properly. MATH!!!

[@VRietyGamer]

AB needs to be equal to AC in order for BB* to be equal with CC*. We know this isn't the case otherwise M would fall on the angle bisector of A. Carlo is assuming the triangle is isosceles in order to prove that the triangle is isosceles. Circular reasoning detected.

[@simplesmilerwastaken]

Yes it is because it lies on the bisector of BC.

[@somewony]

It is though. X is on the right line starting in M. This line is the collection of all points equidistant from B and C. Therefore XB = XC.

[@satyreyes]

That can't be it. X is certainly equidistant from B and C, because X lies on the perpendicular bisector of BC. All points on the perpendicular bisector of a segment are equidistant from the segment's endpoints.

[@satyreyes]

GamerB312 That doesn't follow at all. The line AX has nothing to do with whether XB = XC; you can prove that XB = XC using nothing but the given that MX is the perpendicular bisector of BC.

[@ilv1]

GamerB312 Dude... if you draw a line straight up in the MIDDLE of a horizontal line... All points on that line are the same distance to the edges because it's like you have the bisector of an isosceles triangle which is the same as the median line and other lines. XB = XC... If you say otherwise you'd better go back to 1st grade geometry.

[@michaelgeiss741]

Thanks for the fun trick question! The poorly drawn diagram shows both AB*>AB and AC*>AC, which is not possible. Instead, if AB*>AB then AC*

[@ikasu00]

Holy crow. Now I don't need a ruler.

[@Edgawliet]

Geometry is the art of correct reasoning on incorrect figures -George Polya

[@jaytwocari461]

At 3:36 triangle AXB* is not congruent to AXC* because even if lengths B*X, XC* are equal, angles AB*X and AC*X are equal and angles XAB* and XAC* are equal, there is a difference between angles AXB* and AXC* which is not easy to see at first time because of the weak gap between the bisection and the bisector. That little difference implies that AB is different to AC owing to the non congruence of triangle AXB* and AXC*.

[@MrUMitra]

Well for one thing I know, Congruency for side-angle-side is when the angle is include between the two sides... in the video its just showed us the one angle is 90deg, whereas we need to show that angCXC* = angBXB*

[@mannaggiacristo]

Phytagorean theorem

[@LittlePeng9]

Right angles can have congruence rules a bit relaxed.

[@IhaveWoodforSheep]

en.wikipedia.org/wiki/Congruence_%28geometry%29#Congruence_of_triangles They use RHS, a special case of SSA where it actually does hold true.

[@fouried96]

You're forgetting the congruency rule called RHS

[@Rohishimoto]

IhaveWoodforSheep I call it HL

[@imspidermannomore]

triangles BB*X and CC*X are not congruent. assumption thet they are is false. (you need two pairs of equal sides and the same angle between them to have congruent triangles, any other angle won't do)

[@DrRemiehneppo]

this is actually true cause you got two side equals 2 by 2 and a right angle

[@tim_meister]

imspidermannomore I think this is it

[@imspidermannomore]

***** you're totally right! there's no base to assume that XB*=XC*. i was wrong (i pointed out that triangles are not congruent, but i wasn't correct about the reason why they're not so)

[@newsoupvialt]

XB* and XC* are the same. X is on the angle bisector (the middle between the two big lines) and B* and C* are the shortest distances from X to the two lines. The two big lines are mirror images of each other at the angle bisector. All this means B* and C* are mirror images at the angle bisector. The reason they're not congruent is because if XB* and XC* are the same and B* and C* are mirror images at the angle bisector, B and C would also have to be mirrored the same way for B*B and C*C to be the same.

[@forbidd3nzz]

By using RHS property, you'll be able to prove that BB*X is congruent to CC*X B*X = C*X AC*X = AB*X = 90° BX = CX

[@jameshansen2668]

1:16 is the mistake. M is the locus from B and C, but A's angular bisector would have to pass through M for the sides to be equal. Using this information we can deduce that AB does not equal AC

[@Tsuyara]

+james hansen The point is to find where the proof makes a wrong step, rather than disproving it being an equilateral triangle.

[@Muhammed_English314]

@@Tsuyara Also, nobody assumed X≠M

[@SKO_PL]

i tried it myself and C* came out to be nearer A than C so...

[@hgjfkd12345]

It's AXB* and AXC*, right? They are definitely not congruent.

[@Keepedia99]

iluvpopcorn23 They are. RHS congruence: angleAB*X=angleAC*X=90deg Hypotenuse AX common Side B*X=C*X

[@Keepedia99]

iluvpopcorn23 They are. RHS congruence: angleAB*X=angleAC*X=90deg Hypotenuse AX common Side B*X=C*X

[@tarseeli8619]

iluvpopcorn23 If the line MX could be extended to AX, then they would be congruent. However, that is not the case. Therefore, you are correct that AXB* and AXC* are not congruent. Also, the fact that BX = CX does not prove that BXB* and CXC* are congruent. There is more than one way to get the same length for the hypotenuse. Thus, the problem with the presentation was that it was based on the assumption that the two smaller sides of a right triangle must be the same if the hypotenuse is the same. That is not the case.

[@jarrenbenin7553]

yeah, AB+BB* must be equal for AC+CC* for this to work. well simplified, AB must equal AC since BB* and CC* are congruent. either way, For the two triangles to be equal, AX should be able to overlap MX. Otherwise, the quadrilateral cannot be bisected equally

[@nuthying3156]

+ㅤㅤ no

[@yaptro]

3:50 -- here is the prpblem. AXB* and AXC* ARE congruent, but though either C or B lies outside AC* or AB* correspondingly, we can't conclude that either AB*-BB*=AB or AC*-CC*=AC. So the main idea is that of of the perpendiculars from x falls inside the triangle. It can be easily prooved just buy drawing approptiate triangle (as an counterexample). In general it's harder to proove it but also possible.

[@griffonthomas7869]

This spoof proof can actually be seen through multiple ways, I see as I am reading through the comments. I saw the flaw in this problem when I realized that in a true equilateral triangle, AX and MX should be on top of each other.

[@hakimal-hakim8890]

AB = AB* - BB* but AC = AC* + CC* NOT (AC = AC* - CC*) The drawing is misleading......

[@martinservold]

You're right, it took me about 30 minutes in AutoCAD to figure it out, but you are completely right.

[@MartiniComedian]

If you freeze the image at 03:03 you will notice the mistake. XBB* and XCC* are NOT congruent. Go and look at the rules of congruency. You are all very welcome. What do I win, Numberphile ??? ;)

[@siblinganon66]

the trick is that the angle at a is nearly a right angle. the two triangles axb* and axc* aren't in fact congruent but are close.

[@Neterskian]

Your sketch is imprecise, thus driving us towards assuming that the points are in this order: A - B - B* A - C - C* while actually it's: A - B - B* A - C* - C (or A - B* - B A - C - C*). Assuming that gives us: AB* - BB* = AB AC* - CC* = AC which is incorrect. From the actual point order we get: AB* - BB* = AB AC - CC* = AC* (or AB - BB* = AB* AC* - CC* = AC). Varying sketches due to different arbitrary triangles (their angles, to be specific), whether X exists or not are also different cases we probably should consider, but this is probably what you were going for.

[@hongkim3662]

To go a step further, could you prove that this configuration is wrong? If you try by multiple examples you will see that this diagram is wrong, but saying why what you said has to be true is a harder challenge. I used a proof by contradiction.

[@Neterskian]

Hong Kim You sound like my geometry teacher. I guess you would consider the different types of triangles (by angles); obtuse, acute, right, and then prove that for every case. Contradiction sounds shorter, though.

[@hongkim3662]

Neterskian I actually was a geometry teacher at one point of my life haha :P If you go by the presenter and say the green triangles are indeed congruent, then that will imply that the angles B*XA and C*XA are congruent. But you can disprove this using other properties of the diagram. Hence the contradiction.

[@tibschris]

I think I know what's going on intuitively, but can't be arsed to prove it. This diagram is deliberately distorted, and any tiny distortion will cause some projections to be wildly off-specifically, with point X. Once you draw the lines from X that create B* and C*, one of these two will points be _interior_ to B or C. In this diagram, they're both shown exterior to B and C which I think is impossible. It is true that AB* and AC* have the same length, but one of these will be true in a scalene starting case: AC > AC* _or_ AB > AB*, depending on which of AB or AC is longer. This means that either BB* or CC* will have, for the purposes of the equations in 3:47, a _negative_ length, so the equations don't work out quite the same way. BB* = -CC* and so AB = AC - BB* so AB and AC are not equal.

[@Serdar54321]

I think perpendicular bisector and angle bisector wouldn't intersect outside the triangle (I mean the right side of BC) (only their extensions would somewhere else)

[@Serdar54321]

or it could be that XC* or XB* would fall inside triangle so "AB*-BB*=AB" or AC*-CC*=AC is wrong

[@Grassmpl]

Not quite. They could still intersect outside

[@fonaimartin98]

They will always intersect outside, expecially on the circumscrived circle.

[@ReidarWasenius]

The fooling step is at 3'25". There is no reason to assume that the points C* and B* are at equal distance from A.

[@isithardtobevegan53]

+Reidar Wasenius They are at equal distance from A but the CC* is not equal to BB*. That is the thing.

[@TechDude351]

he's saying that the angle bisector splits one side from the other equally, even the the midpoint M is the direct middle of B and C.

[@Qwertinator212]

This

[@SuperDewies]

Na its congruent due to RHS u have B*X=C*X the right angle and AX=AX

[@SuperDewies]

Na its congruent due to RHS u have B*X=C*X the right angle and AX=AX

[@tynoArcher]

Numberphile There's a miss conception there, being AC*=AB* does not mean AB=AC nor does it mean BB*=CC*, it just means that sum AB+BB*=AC+AC*, AB can be different from AC (aswell as BB* be different rom CC*) and the sum can still be the same.So this hipotesis is wrong, you cant say that all triangles have the same length size, you can say however that from any triangle you are able to make an equilateral triangle.

[@BintonGaming]

Upon first glance, this seems legit. I've rewatched this several times listening very closely to everything, and I realized something: If a triangle is equilateral, meaning it has all equal sides, that implies that same triangle also has equal ANGLES. This means that the angle bisector of angle A for the triangle you drew must also be the perpendicular bisector for line BC. That is clearly not the case in your example. But all other things check out, so I'm confused.

[@NoriMori1992]

BintonGaming 3:21. Those triangles are not congruent, so anything derived from the assumption that they are is invalid.

[@Athakaspen]

+NoriMori Actually, Those Are Congruent. The Green 3:30 Ones Aren't, Though.

[@NoriMori1992]

+davidthewalker I was referring to the green ones in the first place. I linked to 3:21 for context.

[@Athakaspen]

+NoriMori oh, ok I get it now

[@mdk2able]

+BintonGaming your right, if the triangle was an equilateral the angle bisector would constantly be intersecting with the midpoint of BC, and thats how he proved it by deliberately screwing up the drawing and making them intersect

[@Deathranger999]

I see the flaw. At 3:30, you don't know that angB*XA = angC*XA, thus the two triangles are not necessarily congruent.

[@briankuhns9769]

That step is fine, he is using angles AC*X and AB*X. The problem is that the perpendicular bisector and the angular bisector would intersect above the triangle, not below it as in his diagram. If you try to reconstruct it with the intersection above the triangle it all falls apart.

[@ankitagarwal7896]

Brian Kuhns Kieran Kaempen So like it turns out that one of the lines, either XC* or XB* has to intersect Line AB or AC resepectively, before XC or XB, respectively, so in that case, the entire simulation falls apart.

[@briankuhns9769]

Yes I think that is what I was trying to say, I wish I could insert a google drawing into this to show you what I mean.

[@ankitagarwal7896]

Same, I used Cabri in order to illustrate this

[@karlmuster263]

They are congruent. They have 2 sides equal and they are right, so the Pythagorean Theorem gives you the same length for the third side.

[@joshuayoudontneedtoknow9559]

The issue is that BB*X and CC*X are not congruent because two equal sides with an equal angle that is not where the two lines connect could result in a line of varying length on the other side. Side Side Angle does not mean congruence. It has to be Side Angle Side.

[@joshuayoudontneedtoknow9559]

Indeed, suppose they are congruent, which they may be due to the fact that they are right triangles. However, just stating that triangles AB*X and AC*X are congruent because they appear to be mirror images of one another is not sufficient. On re watching the video, I'm convinced that the trick is at this point. So many lines are drawn that it is hard to tell if he is correct or not (which he can't be).

[@ChongFrisbee]

You should do a find the bogus step in the astounding -1/12 video. That would be awesome!

[@talktothehand1212]

Well if that bogus step is defining zeta (-1) as the sum of all integers than there's the step as that definition is only valid such that the real part of s is greater than 1. Other than that it's completely valid as zeta(-1) does indeed equal -1/12

[@numberphile]

Fábio Reale bit.ly/TonyResponse

[@NNOTM]

Numberphile Link doesn't work, at least for me

[@BaryLevi]

Fábio Reale please learn analytic continuation before claiming to have ripped a high-level mathematics theory

[@talktothehand1212]

Gregery Barton that proof surely didn't work like that nor do the other means at arriving at -1/12

[@qwertyuiopazsd4253]

Trick question. There is no mistake in the proof, all triangles are equilateral.

[@BlizzyFoxTF]

Guys, the thing is that angle B* and angle C* is not necessary be 90 degrees. It can be any angle and that still C*X = B*X. We are duped all because of 1:13. If its not 90 degrees, both triangles are still congruent due to AAS. However, if the angle is not 90 degrees, CC*X not congruent to BB*X, and 2:19 is wrong.

[@Lelentos]

4:06 is the bogus step.

[@artbyfreddiexx]

yep, he's proven that BB* and CC* are equal, so that's like saying 7 - 2 = 5 6 - 2 = 4 therefore they are equal which they clearly aren't

[@sppw463]

+Freddie Guthrie This here, is the correct answer

[@Jivvi]

+Freddie Guthrie I think it's more like saying 7-2=5 7+2=9 Therefore 9=5

[@somayajulapadmavathi9016]

+Jivan Scarano Well said.

[@cecai1a]

The mistake comes from the approximation of the drawing which puts both B* and C* outside the ABC triangle. They are actually always one inside (closer to A) and one outside the triangle with CC*=BB*=|(AB-AC)/2|.

[@prakhar9998]

The point C* will lie in between AC(AC*

[@RedInferno112]

Yeah I see where he went wrong, XC*C is NOT necessarily right-angled. Great fun, btw I have to work out what the square root of z is where z= -3+4i, and my revision books don't tell me how does anyone know how to do it?

[@RedInferno112]

***** It's not though :P

[@kingrobrob]

Let a+bi = root(z) (a+bi)^2 = -3+4i Expand and equate coefficients to find a and b.

[@RedInferno112]

kingrobrob Great thanks so much! I started my further maths course 5 weeks late so had some catching up to do xD

[@ok64234]

kingrobrob sqrt(-3+4i) = 2+I or -2-i

[@Kiko078168]

For any complex z, rational powers of z are given by: z^(m/n) = |z|^(m/n) * e^((i*m/n)(Arg(z)+2*pi*k)) for k = 0,1,2,...,n-1 where |z| is the absolute value of z, and Arg(z) is the principal value of the argument.

[@JeekayTenn]

All Triangles are Equilateral - Numberphile AB*X =/= AC*X This is because Line AX is not in the exact center of the triangle.

[@DrRemiehneppo]

they are in fact congruent, all the initial proofs are right, only the final calculus is wrong

[@fractalground]

This is correct. The green triangles are not congruent. This step is skimmed over and we are told that they are congruent however listening to what he says you can see there is no proof to this assumption. Furthermore, you can see that they are not congruent. Taking the intersection of line AX with line BC as point D, you can see that the two large triangles are split into three triangles each: ABD, BDX and BB*X on the left and ACD, CDX and CC*X on the right. BB*X and CC*X are equal, however you can see that CDX is larger than BDX. CMX is equal to BMX, but the line DX is slightly too far left. The same goes for ABD and ACD.

[@elliottmcollins]

DrRemiehneppo What do you mean "final calculus"? If you see something he's done wrong, feel free to point it out.

[@forbidd3nzz]

What do you mean by AX is not the middle point? First of all, AX is not a point. Secondly, AX is the angle bisector of BAC. So, XAB is equal to XAC.

[@utl94]

forbidd3nzz In order for AB=AC, the line AX must go through point M, which it doesn't.

[@Bratjuuc]

Every time I tried to draw it I noticed that AC always bigger than AC* , if AC>AB. Same goes for AB if AC

[@mavcsquared]

Bratjuuc but you still haven't shown why this happens

[@viditparab2851]

i drew it and one foot of the the perpendicular from x always ends up inside the triangle .... i.e AC* is shorter than AC

[@MrKenkron]

+vidit parab same

[@viditparab2851]

yeah

[@creepypurple6956]

+vidit parab I think you used the wrong letters. AC* cam't be short than AC. "Proof?" AC* - AC = CC* whereas AC - AC* = a line shorter than nothing

[@viditparab2851]

you draw and see for urself

[@creepypurple6956]

vidit parab Alright I did. Still got the same thing. What you proposed is leterally impossible.

[@ImaginaryHuman072889]

There's nothing wrong with his reasoning, BUT the error comes from the fact that the point X is not drawn correctly. He drew the angle ever so slightly off so that the placement of X is incorrect. point X actually should be much closer to point M. The point C* should should be in between point A and point C. you can actually prove that its impossible to have both B* and C* outside of the triangle at the same time. one of them must be inside the triangle and one must be outside of the triangle. Angle Side Side IS valid if the angle is a right angle.

[@NEDinACTION]

ImaginaryHuman072889 Drawing things to scale doesn't matter in geometric proofs. He could have done that whole demonstration without drawing anything

[@ImaginaryHuman072889]

Ryan Harper saying scale doesn't matter during a geometric proof is like saying a rounding error doesn't matter during an algebraic proof

[@gregjang5402]

+ImaginaryHuman072889 But Hasn't this been done before? In Proposition 48 of Book 1 of Euclid's _Elements_ , the diagram was deliberately distorted to find the converse of the Pythagorean Theorem.

[@DeathBringer769]

I think how Imaginary Human had no reply to the Euclid retort/example, even 2 years later ;)

[@GourangaPL]

1:50 here's the trick, the image is not valid, because if B* falls beyond B, then C* MUST fall between A and C, ABB* and ACC* are congruent but one of then would be flipped

[@KubrickFR]

The trick is at 3:24 isn't it ?

[@chrisneto]

that stuck out to me as well... they don't even look congruent

[@robertbereza6335]

Yeah, I'm with you on that :)

[@shinigami052]

Correct. Assuming the triangles are congruent inherently assumes sides AB and AC are of equal length. Everything after that is proving the assumption you assumed.

[@Matthewkyle12]

You're right, because in this step they are assuming AB=AC without saying it, and then tracing back using invisible circular reasoning to say that because they are equal, they must be equal.

[@kdudesmyth]

RHS: Right angle hypotenuse side. The right angles at XB*A and XC*A Hypotenuse is the line AX and they both share it and B*X is equal to C*X Also, because AX is a bisector so the two angles there equal so SAS works too.

[@melayadi]

I think I discovered the mistake. Please correct me if I'm wrong. The main mistake is the drawing and assuming that both |AC| < |AC*| and |AB| < |AB*| at the same time. Only one equality can hold true. This can be proven by showing that at most one of the angles ACX and ABX is obtuse. In this video, it's ACX that must be acute not obtuse as the graph tries to deceive us. The video shifts the points X more than its right position in order for this to happen. Here's the formal proof that the angle ACX is obtuse. Let Y by the point of intersection of BC and AX. Call the angle BAX = CAX = alpha, angle ABX = beta, and angle ACX = gamma. Then by sine law for traingle ACX, we have CX/sin(alpha) = AX/sin(gamma). Similarly, BX/sin(alpha) = AX/sin(beta). Since BX = CX (prove it!), we conclude that sin(beta) = sin(gamma). So, either beta = gamma or beta = 180-gamma. But if beta = gamma, the point Y will coincide with M because traingles ABX and ACX would be congruent and then traingles AYB and AYC would be conguent. If that happened, it's obvious that |AC*| < |AC|. Hence, we are left witn the only option beta = 180 - gamma, which implies that both beta and gamma can never be > 90.

[@justinlasker6269]

You are right!

[@oscarheath5507]

Disproof 3:30 Obviously, the perpendicular bisector bisects the opposite angle in the equilateral triangles opposite to it (i.e. MX bisects angle BXC) Therefore, unless ABC is isosceles, AXC ≠ AXB But by congruency, BXB* = CXC* Therefore AXB* ≠ AXC* therefore AXB* is not congruent to AXC* A very sly diagrammatic slight of hand

[@heimdall1973]

SPOILER ALERT, DON'T READ THE REST OF THIS COMMENT IF YOU'D LIKE TO WORK IT OUT YOURSELF! What's wrong is the position of X. Yes, AB*=AC*. Yes, BB*=CC*. But the equation at the end is not right. They are not both minus, in this case it's AB=AB*-BB*, but AC=AC*+CC* (because AB

[@Formulka]

those 90 degree angles at B* and C* are fake

[@satyreyes]

Drawn a little sloppily in this case, sure -- which bears further thought -- but not fake. X is on an angle bisector. The shortest line segments from any point on an angle bisector to the nearest points on the angle's legs are always congruent and perpendicular to the legs.

[@Formulka]

satyreyes if they are both 90 degrees, then the BB* and CC* are not the same length

[@satyreyes]

Formulka Sure they are, by the Pythagorean theorem. Do you have a proof that BB* and CC* can't be the same length? If so, the mistake must have happened before this step.

[@Formulka]

satyreyes if ACB and ABC angles are not equal, CBX and BCX are then the B*BX and C*CX can't be. The blue triangles are then different and only one other side or angle of them can be equal other than the side they share with the equilateral triangle in the middle. (edited for clarity and fixes)

[@satyreyes]

Formulka You are onto something! -- but I don't think it can be what you think it is. XB and XC have *got* to be congruent and perpendicular to AB* and AC*; that's what it means, by definition, for XB and XC to be the distances from X to the angle's legs from the angle bisector. But certainly XB = XC. So those triangles really are congruent. But then, as you say, ACB and ABC must be equal, and they aren't -- so the error has happened before this step. I think it's related to where B* and C* are relative to B and C. One of the starred points should be closer to A than its neighbor is.

[@CarolinaCastroCoelho]

I think the “bogus step” is at 1:46. Though it is true that the angle bissector contains all the points that are at an equal distance from AB and AC, one cannot draw the shortest distance at a 90° angle from AB and AC, it has to be at a 90° angle from AX. The way he drew it, angles AXB* and AXC* are not necessarily equal (they are only equal in triangles that truly are equilateral), so one cannot say that XB* is equal to XC*. Every following step that depends on this equality leads to a wrong conclusion.

[@makedumitrascu]

The reason this demonstration is not correct not the math, but the drowning itself. More exactly, M isn't really the middle of BC. If it would have been correctly drown, X would have been inside the triangle and AB=AB*-BB* and AC*=AC*+CC*. these new relationships tell us nothing about the relation between AB and AC, as they should.

[@mavcsquared]

Mihai Dumitrascu you can find math errors here. And this isn't a proof, just a conjecture, you have to actually say why this is

[@TheAnonymmynona]

i think the problem is that A X B* is not equal (Congruent) to A X C*

[@TheAnonymmynona]

I tryed recreating this in Geogebra and realised that B* and C* are never bouth in or outside the triangle (somties at B and C) but normaly one is inside and one is out side the drawing they made is inacruate

[@SynysterKezia]

So I could be wrong, but this was what I saw the first time through: I get that this is drawn in such a way as to look correct while being wrong, but it seems to me that even if we assume everything in the video is correct up to 3:20, you can just prove that the AB*X and AC*X triangles are not mirror images of one another by examining their internal angles. Since both triangles contain a right angle and angle BAC/2, the remaining angle in each triangle must be B*XC*/2. But the line segment which bisects this angle is MX, not XA, which is the "mirror plane" between the two triangles. This contradiction indicates at least one of the assumptions is wrong. I can see where he fudged the positioning of X and C*, but was there any other witchery I missed?

[@Chavagnatze]

All triangles are the same. We just look at them from weird perspectives and make them look different from one another.

[@Chavagnatze]

12treeGoldfisher A 2D objects exist in 3D. Orthographic projections are possible too. That is what makes this geometric proof possible.

[@Chavagnatze]

12treeGoldfisher What? The paper is a plane that can be in any possible orientation in 3D space. The equilateral triangle is on another plane and is being projected onto the observer's plane. Projection can be perspective or orthographic. In some mathematics disciplines this would involve coordinate transformations, rotations, or just some just geometry.

[@Chavagnatze]

12treeGoldfisher And yet the video says that isn't so. There are no basics.

[@HeraldoS2]

Chavagnatze And yet the video says that if you beilved it you have been scamed.

[@Chavagnatze]

HERALDO SUPREMO 2 You will find that the geometrically erroneous step is saying that AB=AC based on: BB*=CC* AB*-BB*=AB AC*-CC*=AC BB* and CC* can only be made equal with a perspective transformation. You have to move out of basic geometry to do that.

[@benoitg6933]

AXB* isn't a mirror's image of AXC* !

[@IoEstasCedonta]

+Ben G Yes, it is. They're both right triangles, and they have a common hypotenuse (AX) and a common leg (XB* = XC*), so they're congruent by the Pythagorean theorem. The problem is that either B* or C* should be on the original triangle, without the line having to be produced.

[@arnavnarula1534]

The reason the diagram is incorrect is due to the positioning of B* an C*. I used geometric construction techniques (only compass and straightedge) to draw angle and perpendicular bisectors. My results were that B* fell past AB (Same as in video), but surprisingly, C* fell on side AC so C* was between Point A and C. Using this, it was quite clear why AB isn't equal to AC (Using Geometric Techniques). To begin, I'll clear some misconcepts (Note: =~ means congruent): 1) XB =~ XC was CORRECT. To prove it --> Angle BMX =~ Angle CMX = 90 Degrees (Definition of Perpendicular Bisector) && BM =~ CM (Definition of Midpoint) && MX =~ MX (Reflective Property). SAS Conjecture tells us Triangle BMX =~ Triangle CMX so CPCTC tells us XB =~ XC. 2) XB* =~ XC* was CORRECT. To prove it --> Angle B*AX =~ Angle C*AX (Definition of Angle Bisector) && Angle B*XA =~ Angle C*XA (Based of Triangle Sum Conjecture --> Angle B*AX =~ Angle C*AX (Look at second sentence) & Angle AB*X =~ Angle AC*X (Definition of Perpendicular Bisector) so B*XA = 180 - B*AX - AB*X and C*XA = 180 - C*AX - AC*X. Using substitution, B*XA =~ C*XA). Finally, AX =~ AX (Reflexive Property) so ASA Conjecture tells us Triangle AB*X =~ Triangle AC*X and CPCTC tells us XB* =~ XC*. Now, before moving on, let me just state: The proofs in the video were ENTIRELY CORRECT. The trick was that they used an impossible diagram, so in reality, there could never exist such a triangle with those characteristics. In reality, C* would be on side AC, not away from it. Proof #1) Why does the wrong diagram matter: The video showed AC* - CC* = AC and AB* - BB* = AB. In reality, the equations would be AC* + CC* = AC and AB* - BB* = AB. Notice how this makes a difference, even if we replace AC* and AB* with the variable "x" and CC* and BB* with the variable "y", we get: x + y and x - y. By using the wrong diagram, the video made it seem that the two sides could be represented by x + y and x + y, so they were congruent. Since x + y isn't x - y, AB isn't equal to AC. Proof #2 (Visual Proof): The diagram used some sneaky tricks to appear like a possible triangle. If you pause at 2:01, you can see that #1) The angle at C* clearly doesn't look like a right triangle, despite being marked as one. To make it actually look a right triangle, C* should be moved way closer to C. #2) Angle BAX is shorter than Angle CAX even despite being separated by an angle bisector. This makes a difference because X (the point of intersection between the angle bisector and perpendicular bisector) should be way closer to M if the diagram was to scale. This would move C* even closer to Point A. #3) BM is greater than CM. For the same reasons stated before, C* should be closer and now between points A and C. Although what I've said may look like mini differences, the 3 differences build up one upon another, making an incorrect diagram. Summary: C* is in the incorrect spot and should be between Points A and C, making the entire triangle a flaw in itself. If the flaw was corrected, the same proofs show the exact opposite (that AB is not equal to AC).

[@bojandimovski1504]

AB*X IS NOT CONGRUENT TO AC*X. IT'S THE ANGLE BISECTOR THAT'S BEING THEIR COMMON SIDE, NOT THE PERPENDICULAR!

[@tijmenvanderree487]

+Bojan Dimovski CAPS LOCK!

[@izd4]

+duisjfuds dinges WE NEED TO TURN IT OFF!!!!!!

[@RSPikachuAlpha]

+Ernest Izdebski I CAN'T !!!!

[@sherlockian6770]

WHAT'S YOUR POINT?

[@bojandimovski1504]

MY POINT IS THAT SINCE IT'S NOT A PERPENDICULAR BISECTOR BEING THEIR COMMON SIDE, THEY ARE NOT CONGRUENT, AND THEREFORE, IT DISPROVES THE WHOLE "ALL TRIANGLES ARE EQUILATERAL" CLAIM, ALSO, THANKS FOR JOINING THE CAPS TEAM! LET'S ALL YELL OUTSIDE!

[@skakdosmer]

I love this video! Because although it's easy to see that the drawing is deliberately a little distorted, so that for example the angle at C* is obviously not 90°, it is very difficult to find any flaws in the logical arguments.

[@lucascisneros8147]

Lau Bjerno YES! Finally someone got it right

[@vrcristian]

the small triangles are NOT congruent. quite obvious flaw

[@Stefanox36]

03:04. Two triangles are congruent if one has all the sides equal to the corresponding sides of the other triangle (AB, BC and AC are equal to AB, BC and AC), or two sides and the angle BETWEEN them from one triangle (AB, A and AC are equal to AB, A and AC), or two angles and the side BETWEEN them from one triangle, are equal to the other one. (A, AB and B are equal to A, AB and B). In here the 90 degree angle is NOT BETWEEN the supposedly equal sides.

[@anticorncob6]

Having two sides and an excluded angle being congruent actually is enough when it's a right angle. You can use the Pythagorean theorem to prove that the other leg has to be equal to on both triangles, so they're congruent. That's not the mistake.

[@Stefanox36]

@@anticorncob6. You are right. Now I think that when the two sides are not equal, one of the following things happen: B* is between A and B, or C* is between A and C, therefore the whole thing collapses. In this diagram C* is 90 degrees and every other thing is proposedly not on scale.

[@myadyth7048]

Ah, the equilateral triangle.

[@nepalwomenassociation6201]

The error lies in the fact that, if AB>AC then B* lies between A and B and C lies between A and C*. If AC < AB then C* lies between A and C and B lies between A and B*. In either of the case CC* and BB* are equal and so are AC* and AB* but due to position of B* and C* the proof does not hold. If AB=AC then M and X coincide and B* and C* both remain inside or outside AB and AC simultaneously depending upon the angle of A. Since you assumed that both B* and C* remain outside, the result is obvious that AB and AC would be equal. Nor is your proof complete. You had to consider 3 cases: 1) if both B* and C* remain outside AB and AC (In this case triangle would be isosceles with AB and AC equal and angle A being greater than 90 degree.) 2) if both remain inside and (result = AB and AC equal with A less than 90 degree) 3) if one remain inside and one remain outside. (AB and AC are not equal)

[@Raaarws]

At 1:15 when he draws the bisector and explains that is the locus of all the points equidistant from AB and AC he assumes that they form a right angle, what means that alpha + alpha is a right angle, which isn't right

[@davidabbott2742]

Don't stop here!

[@RyanJiang]

The problem is simply that C* and B* cannot both lie outside of the triangle. Just use Simpson's Line.

[@jhboulder]

The issue is at 3:22. He states that there are "more congruent triangles", and proceeds to label AXB* as congruent to AXC*, without proving AB=AC. Because his drawing of the initial triangle is CLOSE to equilateral, it is visually convincing that the two line segments are equal, and because he is moving along at a decent pace, most people are going to just say "yeah, that looks right" to try to keep up with his explanation. If the triangle were drawn with a much more acute angle included, this point would not be very compelling, and people would catch it right away.

[@LedeEleven]

Nope, congruence is implied by Hypotenuse leg theorem.

[@LalitSingh-xh5yq]

Bro he is not showing a magic trick..instead of calculating the pace of his hand movements,try some math..those triangles are congruent coz of hypotenuse leg theorem

[@djtrig6576]

+Lalit Singh bro are you kidding? You do realize that not all triangles are equilateral right? In fact, statistically speaking no triangles are equilateral...

[@LalitSingh-xh5yq]

David O'Connell obviously, they r not.But the reason is not what u said..

[@djtrig6576]

+Lalit Singh wasn't me, just saw this. But you can't correctly prove two things are equal that aren't equal so you're 100% incorrect as well

[@nicolasdelahoz5417]

Point B* , point M and point C* are collinear, hence or B* is inside the triangle or C* is inside the triangle ( Simpson Line)

[@ArmandoXIII]

1:40 to 2:10, what is this? He is totally wrong here. XB* is definitively not equal to XC*, his explanation is a nonexplanation, it would only apply if the perpenticular bisector was in line with d, which isn't.

[@mattjmwmatt]

.

[@umamaheshwaranl8554]

WTF. he is doing correct only. he clearly said the reason

[@ArmandoXIII]

+uma maheshwaran the reason was a non explanation

[@IoEstasCedonta]

+ArmandoXIII That part is right. Any point on d is going to be equidistant from the two lines, and X is on d. Where he went "wrong" is that he drew the point X too far out - one of the triangles he draws should be partially inside the original triangle.

[@KasabianFan44]

What are you on about? What he is doing at that point makes perfect sense.

[@SteamsNightcore]

It's because Angle, Side, Side isn't valid. One of the few things I remember from high school geometry!

[@numberphile]

Right angles triangles have special rules.

[@professionalfire3902]

Steams Nightcore™ Hah. You got served

[@SteamsNightcore]

Matthew Plascencia Well, I would rather be corrected than to continue being incorrect.

[@professionalfire3902]

Yeah I know, I was just kidding.

[@EpicsRandom]

For a right triangle, if the hypotenuses are congruent and one of the legs are congruent, then the triangles are congruent by HL Theorem.

[@danieltraceski2513]

3:51 and following. It is an unwarranted assumption that AB* - BB* = AB, as is that AC*-CC* = AC. In general B* may lie between A and B, and then AB* + BB* = AB. The first equations will only both be true if the triangle really is isosceles around A [or the addition versions can both be true]; if the triangle is not isosceles, then one will be true and the other false. In other words, if the triangle is NOT isosceles around A, then one and only one of the "star" points will lie within the triangle, and other will lie outside the triangle. Then we'll only be able to say something like " AB* - BB* = AB but AC* + CC* = AC " From these two equations, and given the other identities (use substitution or systems of equations, etc), we will find that AB does not equal AC unless BB* and CC* are zero.

[@natnew32]

I feel like this is borderline reportable.

[@natnew32]

how so?

[@raisins7777]

+natnew32 woosh

[@reissecupfilms]

+natnew32 Why?

[@natnew32]

The title & thumbnail are very misleading on the point of the video.

[@eggynack]

+natnew32 How do you figure? I mean, the title is clearly claiming something that's false, so the video is obviously going to present some false proof. I knew basically exactly what I was going to get when I clicked the video.