Thales Proportionality Theorem

[@TJStellmach]

I foresee a lot more comments saying "just use similar triangles," not understanding that we don't know the properties of similar triangles until we prove them (Thales' theorem being one way to do that). Notice that nowhere here do we even use the concepts of angles having _measures_ outside of perpendicularity and parallelism.

[@goodplacetostop2973]

Thank you

[@matchedimpedance]

No. He also uses the idea that the area of a triangle is always 1/2 base * height. We take that for granted but it is not a trivial matter.

[@obnoxious2471]

@@goodplacetostop2973 are you an official account or just an active viewer?

[@goodplacetostop2973]

@@obnoxious2471 An active viewer

[@kumarsaurav8885]

For people having a sudden Cartesian style loss of faith in what they know, here’s a path to this proof from Euclid’s postulates: 1) Define by the area of a rectangle the products of its sides. 2) Show that the area of a parallelogram having the same base as a rectangle between a pair of its parallel sides is the same as that of a rectangle by showing that the two triangles so created are congruent. 3) Extend 2 to any parallelograms between same parallel lines drawn on same bases. 4) Show area of triangle is half the area of parallelogram drawn on its base. 5) Use above to show area of a triangle is half the product of its base and height. We are now in a position to justify Michael’s proof.

[@blackpenredpen]

5:27 pro skater in the making!

[@aaaaaa-rr8xm]

hi

[@MathElite]

That's a super interesting geometry problem. I never thought of proving that

[@andreivila7607]

I can sense an IMO geometry coming :) Hope it actually happens...

[@gigagrzybiarz]

I'd be so happy

[@blazedinfernape886]

Geometry is super hard ;(

[@deepakgoswami7882]

:)

[@mamv2001]

It will be epic

[@garvittiwari11a61]

True, I also hope

[@goodplacetostop2973]

5:26 Good Place To 🛹

[@indubansal9398]

Hi sir big fan of you and mathematics

[@AnkhArcRod]

Neat! I always find it hard to imagine the constructions! In this case, there are obviously other ways to prove the identity...but constructions are mainstay of harder geometry problems!

[@tobiaskorn9633]

I would construct the height of the triangle ABC with respect to the side A. Then I would call the angle between AB and the height Alpha, the angle between AC Beta and the intersection point from the height with DE as F and the intersection point with BC as G. Then I could write following equations: AD = AF / cos (Alpha) AB = AG / cos (Alpha) AC = AF / cos (Beta) AE = AG / cos (Beta) The by use of this equations it will be clear that AD/AB = AF/AG = AC/AE

[@mrhatman675]

Your proof is invalid cause tan cosin and sin actually originate from triangle similarity which is in turn proved by thales theorem

[@lightyagami1752]

To me using similarity makes this trivial. The similarity of ADE and ABC is more "obvious" than Thales so I'm not sure why it isn't natural to start with that. AB/AD = AC/AE (AD+DB)/AD = (AE+EC)/AE 1 + DB/AD = 1 + EC/AE DB/AD = EC/AE AD/BD = AE/EC (QED)

[@diegoalbertoduransalas4806]

Great :)) I just took up an Elementary Geometry book as well, hope to see more of these videos. Thank you :))

[@marouaniAymen]

I really liked the proof, it's more exciting than using similar triangles

[@petersievert6830]

Have the same problem, Michael. Not being that much into geometry, I often present my pupils problems from other fields, that I am more enthusiastic about ;-)

[@kultek]

Thank you. This helps me get a better insight into a lot of geometry proofs. Very elegant.

[@captainsnake8515]

Glad your covering more geometry. I’ve been taking classes in advance combinatorics,and your other videos have already strengthened my algebra and NT, so geo is by far my worst topic. Can’t wait to see more!

[@xicufwm]

1000th like and I'm proud of it

[@farruxhikmatullayev9120]

Thank you for solution. It was very helpful for me.

[@_judge_me_not]

Just draw EF || BD Create parallelogram Prove ADE similar to ABC

[@jackhandma1011]

Really cool. I always use similarity and didn't even know the proof 😂

[@patrickpablo217]

Cool! You can use almost the exact same picture to prove the Triangle Areas Ladder Problem, and then use the Thales Proportionality Theorem itself to prove that the Triangle Areas Ladder Problem solution works for all triangles (right, acute, and obtuse) without needing to prove the cases separately.

[@beegdigit9811]

5:27 vlogger Micheal

[@goodplacetostart9099]

Good Place To Start at 5:27

[@syedrehanfida]

Looking forward for more skating clips :)

[@mohammedelhouari6468]

Fantastic. Congratulations !

[@ПетрКурнев]

The proof is the best of the best 💥

[@stvp68]

I’d like to see you look at Newton’s geometrical calculus proofs

[@ra-hu3lu]

Continue your abstract algebra playlist and do advance abstract algebra

[@noahtaul]

In a lot of these problems it's not clear if you're going for exploration or just for the fastest (or most standard) proof possible. If the latter, you can say that ABC~ADE, so AB/AD=AC/AE, then subtracting 1 (since AB=AD+DB and AC=AE+EC) gives the answer.

[@K1234-j4h]

Similarity of triangles is proved using BPT and congruency(of triangles not congruence modulo).

[@names9769]

How is this man so cool on the skateboard yet still feels like a teacher when he rides it?

[@jacoboribilik3253]

can we always drop a perpendicular from any vertex of a trinagle to the base?

[@laurynastruskauskas6586]

damn i actually want to get better at geometry too, nice to see that there might be some interesting geometry problems comming up, cant wait.

[@ilickcatnip]

In many textbooks in India this is actually called 'basic proportionality theorem' or simply BPT and idk why.

[@xspeedox]

Id derive from the sine theorem. Or similar triangles.

[@jamesfortune243]

Flip horizontally over vertical altitude to A, leaving original triangle there too. Trivial case or wolog as you've shown. Similar triangles ADE' and ABC' by parallel theorems. AD/AE' = (AD+ DB) / (AE' + E'C'). Cross multiply and solving gives desired result after comparison of lengths to original.

[@alejosanchez3309]

This proof is more difficult a¡ than it seems, as it would first require developing a theory about areas, which is very complicated, and the theory about lengths and proportions is much more fundamental.

[@joshuagiusti1280]

Prove or disprove that for all natural number n bigger or equal to 5, 6^n+1 is composite.

[@joshuagiusti1280]

Source: I don't know...

[@sudeepsingh1298]

If n has an odd part(other than 1), the term is composite. As such , we only need to evaluate for n=2^q, where q is a natural no greater than/equal to 3. This is where I am stuck. Care to provide any hints?

[@joshuagiusti1280]

@@sudeepsingh1298 You can use prove by induction. Also notice that if you use n=k+1 (k is a natural number) use Contradiction Proof, and use the form of all primes except 2 and 3 to prove that is impossible. That is, use the fact that all primes (other than 2 and 3) they're of the form 6m+1 or 6m-1 right after the Induction Hypothesis, but you'll get a contradiction.

[@sudeepsingh1298]

@@joshuagiusti1280 How have you got a contradiction using induction? Can you provide the solution?

[@TechToppers]

Hmm... For odd n, it can be factorized easily. For even *n* we have to do some work...

[@prathmeshraut1616]

Nice video

[@MrRyanroberson1]

Definitely elegant

[@cernejr]

Wow. Thales lived around 600BC. What would the Ancient Greeks think of us today?

[@sharpnova2]

they would see us as gods until they properly accustomed themselves to modern tech. then they would be similar to us

[@heferh4320]

isnt thales theorem more of paralels and secant lines? the segments between these paralels are proportional

[@curtiswfranks]

How did you draw such straight lines?

[@rendyadinata5534]

Here is a much faster way (I think), since triangle ADE and ABC are similar , AD/(AD+DB)= AE/(AE+EC) => DB/AD +1= EC/AE+1 => AD/BD= AE/EC.

[@TechToppers]

The ratios you are setting up requires Thales theorem for a proof.

[@joshuagiusti1280]

Are there infinitely many primes of the form n^2+1 where n is a natural number?

[@aadfg0]

This is an open problem: en.wikipedia.org/wiki/Landau%27s_problems

[@sharpmind2869]

This is the same proof as given in my 10th class maths book , nice video though. 😊😊

[@hdt1948]

is it possible to prove it without using the concept of area? I think it would be a stronger proof.

[@opresopre]

Neat proof!

[@АндрейВасин-в7я]

Even though the proof is really elegant, it kinda have the same problem as most of the proofs of this simple elementary geometry facts. For this proof to be valid, you need to prove that area of rectangle is m*n. The standard proof that I know is dividing rectangle in small squares and taking limit for irrational sides. The standard proof of Thales Theorem is pretty much the same: you divide one segments into many smaller segments and take limit for irrational lengths. So this proof just uses the fact of similar complexity as it tries to prove.

[@tomatrix7525]

Guys, similar triangles is a logical fallacy because we don’t know similar triangles at this level yet. It’s like using running to help you walk.

[@lucho2868]

glad to see you found the way out the forest

[@goodplacetostop2973]

HOMEWORK : Let S be the set of integers that represent the number of intersections of some four distinct lines in the plane. List the elements of S in ascending order. SOURCE : 2010 Lexington Mathematical Tournament

[@K1234-j4h]

Maximum no. of points of intersection will be 4(4-1)/2=6 So S can contain only integers between 0 and 6 both inclusive 0 points of intersection when all lines are parallel. 1 when all lines are coincident. 2 ( I think its not possible. If anyone knows then do tell) 3 when three of the lines are parallel and fourth intersects all three parallel ones 4 when there are two pairs of parallel lines. 5 when two lines are parallel and remaining two are not parallel to any other. 6 is maximum possible. So S={0,1,3,4,5,6}

[@goodplacetostop2973]

@@K1234-j4h SOLUTION 0: all four lines parallel; 1. all four lines passing through the same point; 3: three lines parallel, and the fourth passing through all of them (as a transversal); 4: two pairs of parallel lines, forming a parallelogram; 5: two parallel lines with two transversals that intersect off the parallel lines; 6: four lines, no two parallel, no three passing through the same point. To see why we can’t have exactly two intersections, start by drawing two intersecting lines. If we draw a third line, it cannot intersect both lines in different places, or else we get 3 intersections, so it must either be parallel to one of the lines, or pass through the point of intersection of the first two. In the first case, we have two parallel lines and a transversal, but the fourth line will have to intersect one of the lines somewhere, giving more than 2 intersections. In the second case, the fourth line cannot pass through the point of intersection of the first three lines, or else we only have 1 intersection, but it can be parallel to at most one of the three lines, so we’ll get at least 2 more intersections with the first three, for a total of at least 3 intersections. Either way, it’s impossible to get exactly two, giving our answer.

[@aadfg0]

@@K1234-j4h 0, 1, 3, 4, 5, 6 are easy to construct, but you cannot get 2. Define a~b if a, b are parallel. Since 1 equivalence class means 0 intersections and ≥3 equivalence classes means ≥3(3-1)/2 = 3 intersections, we must have 2 classes with sizes 3 and 1, giving 3 intersections, or 2 and 2, giving 4 intersections.

[@shashwatranjan7437]

Demn , in India you learn the proof in 10 grade (15yr olds) , didnt expected that u will make a video on it

[@xy9439]

That pronunciation of "Thales", though

[@nikitakipriyanov7260]

@@ManwithaCat In Russia we used to call him Фалес, which sounds, well, first sound is clearly like "F". Never knew its name starts with theta Θ. It's a pity teachers don't teach that stuff. "Thales" was so unusual to me so I even didn't recognized him until I saw a statement of the theorem.

[@Myrus_MBG]

Man with a Cat I personally pronounce it /tʰɑ.ləs/ just b/c I’m a little too obsessed with ancient languages in the PIE family. I also say /py.tʰa.ɡo.ɾas/ though, admittedly just to be slightly annoying to my teacher.

[@Lukasz.Skowron]

What if angle at A is obtuse? Isn't that proof only valid if A is acute?

[@HaiNguyen-qx3db]

Nah. The proof is still valid if A was acute.

[@Lukasz.Skowron]

@@HaiNguyen-qx3db Yes, but the altitudes would drop outside ABC. I know it works even though. But I think the proof would look slightly differently with different points etc.

[@theguythatmakesyoumad3834]

Why not simply use similar triangles?

[@lc1777]

Because similar triangles are derifed from Thales theorem 😃

[@kriswillems5661]

Sometimes it's difficult for an engineer to see what's going on in a mathematician's mind. If you scale up something the scale factor is equal in all directions and all measurement also scale up with the same factor. I am surprised this even needs proof.

[@TJStellmach]

This proof (or the equivalent proof about similar triangles) is how we know that this construction with parallel lines represents the kind of scaling you describe, and isn't, for example, some kind of optical illusion.

[@segmentsAndCurves]

@Anton Melnyk I think similar triangles are proven by Thales Proportionality Theorem, not the other way around.

[@HaiNguyen-qx3db]

@@segmentsAndCurves And that is how we learn about criteria of similar triangles at grade 8 here

[@segmentsAndCurves]

@Anton Melnyk Which axioms were you mention.

[@Billlakeview]

To me, the issue is about a broader problem in mathematics....to assume the least via postulates and be able to prove everything while moving from the most basic to the more complex. For example, in most high school geometry courses in the USA, parallelism is covered before congruent triangles requiring 2 postulates , one to conclude lines parallel and another to get conclusions given parallel lines. However, if you do congruent triangles first, you only need SAS to eventually prove the theorem that alternate interior angles congruent implies parallelism. BTW, SAS is enough to prove most of the other congruent triangles results like ASA and SSS. Doing Thales Prop theorem before the similarity theorems means you don’t have to make postulates on similarity. But if you are not trying to build a logical system based on the least number of assumptions, the argument can be made that it is more important to make more assumptions as long as they are based on generalizing experimental data. Sounds a lot like the argument of science and applied math vs pure mathematics....

[@lc1777]

Geometry is ridiculously difficult, how do you guess the constructions is beyond me

[@vanandreyambalong628]

First comment that didn't time travelled

[@luna9200]

Nice!

[@sharpnova2]

I'm a little concerned with your use of the triangle area formula here.

[@edan6758]

In my secondary textbook they said:” Admitted, No prove” :))

[@segmentsAndCurves]

Relatable.

[@ApresSavant]

Why such an unnecessarily convoluted route? If we know ABC, we know ADE is similar. So the ratio of AB/AD would be AC/AE too. Obviously also AB=AD+DB, and AC=AE+EC. Substituting in each we quickly see that (DB/AD +1) = (EC/AE +1), and reducing (cancelling the two ones and inverting), we have the desired ratio.

[@TechToppers]

The proof that those ratios of similar triangles are equal comes from Thales theorem. He did it like to avoid circular reasoning.

[@ApresSavant]

But that in itself is circular reasoning, as the method he used required the area calculation, which is ultimately based upon squaring the triangle with a mated rotated similar triangle. The reality is that the similar triangles argument can be derived by so many other routes, and stands alone with results like Thales as corollaries..

[@lc1777]

@@ApresSavant can't you derive the area by concept of congruent triangles?

[@TechToppers]

@@ApresSavant Area of a triangle comes from well known theorems and it really doesn't use similar triangles. If I'm wrong, do correct me mate.

[@egillandersson1780]

It is always a good idea to start with hippies 🧝‍♂️🧝‍♀️

[@taufiqutomo]

What's with these recently declassified videos? There is (or at least was) a secret mathematical society out there.

[@ezequielangelucci1263]

hi michael, in a lot of videos you finish it too fast and i cant see the board because the video ends with you in front of it can you in future videos wait one second to finish the video so i can stop and see? srry for my english

[@ezequielangelucci1263]

great video : D

[@ezequielangelucci1263]

en español: ¿podrías dejar un segundo con el pizarrón en la pantalla sin que te pongas en frente así se ve bien claro todo antes de que finalices el video? muchas veces no llego a leer lo que aparece y necesito ayuda visual ya que mi ingles no es muy bueno que digamos xD.

[@jesusthroughmary]

I would have labeled it DB to be consistent

[@gordonstull1962]

That is cool... I can see something here that I didn't realize before, in that the pulling-force,(to pull-straight) can be plotted in a crystalline structure as a balanced closed-looped particle-flow. Is there a formula that could show a triangle building itself... like in the case of the RNA & DNA molecule? Would that even be possible? Love your videos...

[@CM63_France]

Hi, For fun: 2 "ok, great", 1 "so let's may be go ahead and".

[@Jimothy-723]

cool. now for your next trick: solve the Thales Theorem Arithmeticaly, and not Algibraically.

[@srajanverma9064]

We learn this in grade 10th ( in India)

[@segmentsAndCurves]

We learn this in grade 8th (in Vietnam)

[@dariuschitu3254]

We learn this in 6th grade (Romania)

[@srajanverma9064]

If that is the case, why is this man, who generally deals with college physics or higher secondary maths or olympiads...teaching this!!

[@dariuschitu3254]

@@srajanverma9064 I think because it relates to much more complex topics (if you're using The Fundamental Theorem of Similarity), like Ceva's Th., Menelaus's Th., Van Aubel's Th., Theorem of the transversal, Steiner's Th., Carnot's Lemma, Stewart's Lemma etc.

[@segmentsAndCurves]

@@dariuschitu3254 Indeed.

[@Stelios2711]

Nice, but the idea behind that proof was not motivated.

[@markbracegirdle7110]

Can't you just say that △ADE is similar to △ABC? Then it follows immediately.

[@nickjackson3061]

Me who seen this and studied this proof in class 10th:- "Too easy"

[@lc1777]

Yeah yeah , you "read" this proof, you didn't derive it yourself, ofcourse if you read it , it will be easy

[@avyuktsinghraghuvanshi2498]

pove that : 123456789123456789.......123456789 is neither a perfect square or a cube

[@LearnwithTejeshwar]

This is our class 10th board exam most important question.

[@HaiNguyen-qx3db]

🤔

[@AnmolTheMathSailor]

Hey sir big fan can I accept a heart from you 🙂

[@srijanbhowmick9570]

you can only accept the heart when he gives you the heart 🤣

[@AnmolTheMathSailor]

@@srijanbhowmick9570 yup but he can even reply on 😪

[@ARKGAMING]

Should be quite easy with similar triangles from the corresponding angles based on the parrallel lines

[@TJStellmach]

Okay, then you need to prove at some point that that's a property of similar triangles. Thales' theorem is equivalent to that.

[@dqrk0]

u cant prove any of the rules for similar triangles without using thales theorem, so u cant use it

[@sayanjitb]

it can also be proved by showing the similarity between that small upper triangle and the big one.

[@timetraveller2818]

hello present human you might notice that the publish date of this video is less than the time i commented so i have proven that i am time traveling Well ,you see there's people that think that this video was private Don't believe them!

[@K1234-j4h]

Ya it's just that this video was hidden and obviously time travelling is not possible🙄😒😒 Well I am commenting too before it was published

[@timetraveller2818]

@@K1234-j4h bruh it's a joke

[@reshmikuntichandra4535]

@@K1234-j4h Though it's a joke,time travel is hypothetically possible

[@timetraveller2818]

@@reshmikuntichandra4535 yea time travelling to the future is possible.

[@anishgupta3992]

the triangles are similar so AD/(AD + DB) = AE/(AE+EC). Multiply denominators and cancel to find AD*EC = BD*AE. Divide both sides by BD*EC to find AD/BD = AE/AC as required. Not sure why this video is so long.

[@TJStellmach]

The facts you're using about similar triangles are also a theorem that ultimately requires proof. Thales' theorem is essentially equivalent to that.

[@anishgupta3992]

@@TJStellmach Thanks, though it is not at all clear to me beforehand where Thales' theorem fits in the order in which we develop elementary geometry. I'd have liked the uploader to mention this.

[@trueriver1950]

Too complicated. Triangles ABC and ADB are similar (three angles, parallel lines). Therefore the corresponding sides are proportional. Therefore the differences in corresponding sides are also proportional.

[@TJStellmach]

The thing we're trying to prove is one of the core _properties_ of similar triangles, so you're engaging in a circular argument by using the properties of similar triangles to prove it.