How To Solve The Hardest Easy Geometry Problem

In the figure, what is the value of angle x? This problem is known as Langley's Adventitious Angles. It is also known as the hardest easy geometry problem because it can be solved by elementary methods but it is notoriously difficult to work out. Can you figure it out? The video presents a solution to this tricky geometry problem.
*At 5:04 I misspoke. I meant BG = BF. We wil shortly prove BG = GF.
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[@MindYourDecisions]

A small correction: at 5:04 I meant to say BG = BF. We wil shortly prove BG = GF.

[@onyebuchiukeoma1551]

Pls why do we have to draw a line while solving this question rather than solving it as it is

[@Ruslan-uv3xb]

@@onyebuchiukeoma1551 because that's how you *can* solve it...

[@justinnicklin4366]

@Vinz Vinz It isn't assumed. Its because triangle BFG is equilateral (all angles are 60 deg, therefore all sides of that triangle are equal) :)

[@rizo1044]

plz someone check my comment lmao its somewhere in the newest

[@charlesrodriguez2176]

@Vinz Vinz I was thinking the same thing - how do we know they're equal?! I didn't understand that statement.

[@leif1075]

And how does one come to think of drawing that magical initial isosceles triangle in the first place?

[@nyanbrox5418]

the answer is you either don't because that specific method only works in this exact scenario or you learn big people maths and try to form a general solution for whatever angles EBF, CBE, BCF & ECF are. you are gonna need some trigonometry! the only reason it works in this case is cause the entire question is done in a giant isosceles :P

[@benjamingardner3314]

You don't need trig, plain old geometry works fine, just know 180° is a triangle, 360° is a square, and start solving.

[@erickjuma7643]

Hahahah that's what I was embarrassed to ask...

[@paolo_castellani]

this isnt the easiest way

[@playerunknown7211]

Experienced Geometry Nerds.

[@redlt5292]

did it in 5 seconds , I just had to grab my protractor

[@theworld2704]

Lmao

[@secretunknown2782]

Lmao

[@thefuturegamer5159]

Lmao

[@FireStormOOO_]

Got a quick draw holster for your protractor do ya?

[@seemakochar5627]

Lmao

[@maximedition8278]

How did no one not notice that there is ABCEFG but no D?

[@notche3358]

thats not the point at all

[@hashirshaikh510]

Lol doesnt matter

[@IanNewYashaTheFinalAct]

Point D is the intersection of BE & CF, duh! :P

[@zombiestrange6475]

Because this question is a girl.

[@dewthevaporeon5977]

You'll get the D later

[@Tacticaviator7]

Damn, after watching like 10 of these videos I finally solved this one myself, I'm so proud of myself for solving something that a Chinese 5 year old would do in seconds.

[@armaansharma8349]

Dude I swear this is the only problem I did by myself. Not too hard, but he made it sound kinda complicated. I just did some basic stuff.

[@BiIIal]

I've do these types of questions on a daily basis and seem to have no problem whatsoever.

[@epikherolol8189]

@@armaansharma8349 it's a class 9 question so I solved it, I'm too in class 9

[@secretunknown2782]

I am preparing for Olympiad and I also did it myself

[@enrilenaminecraft3680]

*Math video exists* Indian dudes: c'mon let's brag here.

[@somapal3654]

This problem was not so hard.But you made it look really complicated

[@randomdude9135]

I assume u have passed Jee Adv by now🤣

[@masda2222]

It is easy to talk Try to solve it

[@rapizer3427]

he is over complicated it its easy, I solved this question in 3 mins when I was jus twelve years old :| edit: now that I'm older, I realize how idiotic and braggy that sounds XD

[@legionSpat]

bundle maarte reh... tere jaise feku bahut dekhe hain... exam mein saari hawa nikal jaati hai...

[@spiderjerusalem4009]

Hillarious that people here have been complaining about how the OP's attempt made it hard, but when they try to elaborate it 𝗴𝗲𝗼𝗺𝗲𝘁𝗿𝗶𝗰𝗮𝗹𝗹𝘆, they either somehow suddenly know the measure of one of the unsolvable angles or merely straight ahead solve it trigonometrically What's funnier is when people say, "All you need to know is that the sum of all triangle's angles equal to 180° while quadrangle's equal to 360°" "Verticle angles are equal to each other, so you get that left = right = 70°" but there are LITERALLY obvious common facts that people wouldn't feel the need to bring up. How ABOUT you tell us how to solve it by doing your so called "oNLy cOunTiNg, nO dRAwiNg" method. I've really really wanted to see that for like 4 years already, or since the video was posted. The problem with the "merely counting" method is that it would always end up getting 180°=180° as the final count and result. You'd always get stuck on that. You wouldn't get anywhere or even move forward. if those "obvious common facts" were the mere needed things to know of, then such question wouldn't have come out in the first place. Stop this hipocrisy and bs once and for all. Once again, you either solved this trigonometrically OR just wanted attention and reactions by commenting another "yOu pOinTleSsLy mAdE iT hArD" due to what the majority of the comment section have been saying.

[@LucidSpaceDog]

I've forgotten a lot the stuff I was taught about geometry around 40 years ago. So I just used what I remembered. The fact that all the angles of a Triangle add up to 180°, lines that cross add up to 360°, angles coming off a straight line add to 180° and the angles of a Quadrilateral add to 360°. Then I went looking for all of the above shapes and situations. Call the point where the lines BE & FC cross G We know ∠ABC = 80° & ∠ACB = 80° ∴ ∠ BAC = 20° We also know that ∠ BGC = 70° Now consider Δ BGC We know ∠ BGC is 70° Now consider Δ BCE We know ∠ BEC = 40° Now consider Δ ACF We know ∠ AFC = 130° Now consider Δ BCF We Know ∠ BFC = 50° Now consider Δ ABE We Know ∠ AEB = 140° Now consider Δ CEG We Know ∠ CGE = 110° Now consider Δ BFG We Know ∠ BGF = 110° Because we know 3 of the angles at the crossing point of the line at point G and they add to 360° ∴ ∠ EGF = 70° Then I got stuck. I couldn't make any more of the shapes I remembered about. I Knew I'd got 110° to share between the angles BFE and CEF in the Quadrilateral BCEF And 160° in the Triangle AEF So I looked at the solution. You cheating bastard :) You add extra lines and angles - Also I'd completely forgotten about the properties of Isosceles triangles :) Thanks though. It was nice to drag some things I was taught so long ago out of the back of my head.

[@AvoidsPikes-]

😂 I said the same thing about isosceles triangles. That got me!

[@Diogenes323]

I stared at the thumbnail for 5 minutes and sold it. then click on the video.

[@knucklepunch5435]

What?

[@kronologie]

solved*

[@indrajitbagchi7313]

good work......"Will Hunting"

[@panology6985]

same I didn't keep track of how long though I think it was 2-3

[@knightowl2407]

Tory Berry SAME! Although I got 50... I came so close... although I haven't taken any geometry. But I do know the three angels = 180 and what he said in the video. Maybe I should have checked my work.

[@dalevillanca6990]

i started watching 1 vid. now im addicted to his voice! omg

[@belawessollek3570]

+1

[@fayedingle6675]

Ok I had no idea you were allowed to add in random lines! How on earth are your supposed to know to do that?!

[@SingNostalgiaWithAmogh]

It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°. Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE. Lastly, use inscribed angle theorem : Angle BEF = 0.5 × BGF = 0.5 × 60° = 30° Done !!! ✌️✌️☺️☺️☺️

[@mmmecho]

Much simpler solving with supplementary angles.

[@cs8833]

please explain!

[@anandk9220]

@@cs8833 They can only comment, they can't explain. [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊

[@SingNostalgiaWithAmogh]

​@@cs8833 ✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️

[@SingNostalgiaWithAmogh]

​@@cs8833 Shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️

[@ramakrishnamansoku]

I think it is 40 degrees because in triangle CBO we have 60 +50=110 Then180-110=70 Since triangle is issocless according vertical opposite angles one of its angle is 70 degrees since it is a isosceles triangle it becomes 70+70+x=180 then x will be 40 degrees....I think it is my opinion so if it's wrong sorry....

[@kazishahjalal6852]

Yes I tried it repeatedly and got 40 degrees somehow

[@anandk9220]

@@kazishahjalal6852 Here's the most logical and easiest way... [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊

[@SingNostalgiaWithAmogh]

​@@kazishahjalal6852 ✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️

[@Scarabola]

4:55 "we'll consider this triangle BFG" Doomguy approves

[@mrslasher9340]

Dude thank you I was having a headache about this and you made it ten times more enjoyable

[@RAJA486848]

kzbin.info/www/bejne/gKvddqR7eNSXpJo

[@deepakmanwaniappdev]

i can explain it to u only using angle sum property and without any construction 🙂🙂

[@Crazy77772]

@@deepakmanwaniappdev bro please tell me

[@anandk9220]

@@Crazy77772 [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊

[@gideon_kang1]

huh? I'm not sure how everyone in the comments don't realise that you can figure out the method in a few seconds. you don't have to draw anything at all! this question doesn't need to use ANY isosceles triangle logic at ALL.

[@mathsplus01]

I really enjoyed this video. Its in some way a perfect presentation and solution all in one

I didnt thought about drawing new lines. Time has passed since elementary chool

[@androidkenobi]

wow, I did geometry in high school. isn't memorization of theorems too strange and boring for tiny kids? well... i guess high schoolers have their own issues

[@nellvincervantes3223]

Answer this question PreshTalwaker An elliptical plate initially rests on a horizontal surface at position where its major axis, 4 m, is in vertical position and its minor axis, 2 m, is in horizontal position. Determine the angular velocity of the plate after it is released from rest, at position when its major axis is in horizontal position and its minor axis is in vertical position.

[@ExaltedDuck]

Easier and less messy solution: Construct a line through C which is parallel to AB. This will be 20 degrees below AC. Then an extension of EF to the new line will create a new triangle. Start solving angles using 180 degree sums of triangle vertices and congruency.360 degree sums of vertical angles. The new triangle will be shown to be isoceles and every angle in the diagram can be resolved.

[@williamrussell873]

That's the elegant solution and the one I found too. And it proves Michael Parks' claim that angle EFA is 50 degrees. Oftentimes these problems give rise to way over-complicated solutions.

[@Yellow9Ball]

Just tried it your way and still can't get it. What is the "This will be 20 degreees below AC." mean? Diagram or/and calculations would be great to see if you have the time. Thanks!

[@ExaltedDuck]

I'll try to explain it. Basically, it's an extrapolation of the vertical angle theorem. If you pause the video at around 30 seconds, I'll try to build off Presh's hard work. We can go about either way (construct parallel, prove 20 degrees or construct 20 degree, prove parallel). Lemmas: Angle ABC is 80 degrees (60+20). Angle ACB is also 80 degrees (50+30). We will construct a new line through C and choose a point Z arbitrarily to the right. We will also extend BC downward and pick an arbitrary C' below C. If we know we have drawn the new line CZ parallel to AB, then angle ZCC' would have to equal angle ABC which we know is 80 degrees Then BCC' = 180 and BCZ = BCC' - C'CZ = 180 - 80 = 100. Finally,, ACZ = BCZ - ACB = 100 - 80 = 20. Alternatively, we can choose to construct line CZ such that it is 20 degrees below AC. Then BCZ = 80+20 = 100 and ACC' = BCC' - BCZ = 180-100 = 80. Since ACC' = ABC, we can conclude that AB is parallel to CZ.

[@2C00L2LIVE]

The new triangle cannot be isosceles, since x will be 60 in that case.

[@ExaltedDuck]

2C00L2LIVE x is 30. the triangle I've proposed building would come out to 80-50-50 and be isoceles..

[@okoroanthony]

This really helped a lot. My teacher told us to come here to get the solution and it worked

[@jaromy03]

I just guessed 30° because is looks similar to the other 30°angle

[@JavierFernandez01]

I think that's how he got his 20° perfect angle to start with anyway. hehe. it's 40!

[@tc1817]

no. for triangle BCA he just added angle B (20+60) + angle C (30+50) = 160 180-160=20 Not too fucking hard if you went to 6th grade.

[@shrirambhandari1463]

I am highly convinced that your solution is quite convincing. The other solutions given here are not that convincing as they are bound to make assumptions.

[@befree984]

You just posted the same problem.. with.. 20° and 10° Like just now..

[@georgemissailidis1504]

Adding CBG made things more complicated. Just use the fact that sum of angles makes 180 degrees, and consider the triangle BCE. You get that angle CEB is 40 degrees. Boom, just like that. Then you can work out all angles easily except x and CFE. So *then* apply your stuff. Easier.

[@andrasnoll2559]

I think I solved it without drawing triangles: so we have a 70° on the middle cause 180° -(50°+60°) Now notice that in the middle all four segments add up to 360° the back to back opposite sided ones are = so 70° - 70° pair and by extension the top and the bottom ones are a 110° - 110° pair {360°-(70°+70°)}\2= (360°-140°)/2=220°/2=110° After that by extension we have 2 more BCE 180°-(30°+110°)=40° and. BCF 180°-(110°+20°)=50° Now we know that 50+CEF+AEF=180 And that 50° and AEF is eachother but mirrored so 50°=AEF (you basically can draw a line through BA line in F point at 90° and get a simetry that also cuts the middle angle exactly in half) that means that the middle is 180°-(2•50°) so 80° Now we have all three angles 70° 80° and x so 180°-70°-80°=x=30° We can fund out all angles inside without drawing lines really.

[@AQWraghd99]

Anyone who thinks they "solved it" in 2 minutes by observation should propably relearn math

[@Slaave]

Wfreestyle why? By glancing at this for 30 seconds I can tell you every intersections angle without a pen. Let's take the intersection between BE And CF and call it G so I can explain this. Triangle BCG needs to add to 180. We have 60 and 50 for 2 corners, this means angle G of BCG is 70. That means in EFG that angle G is also 70 due to it being symetrical. The angle around line segment CF needs to add up to 180, if the G angle of BCG is 70 then the G angle of BGF is 110. If angle G of BFG is 110 and angle B is 20 then angle F is 50. The line segment BA needs to add up to 180, we know the F angle in BFG and FEA are 50 degrees which means GFE's F angle is 80. That means E is 30 as triangle EFG needs to be 180 If we wanna know the rest than angle E of CEG is 40. Line segment CA needs to be 180 so angle E of EFA is 180-30-40 so it's 110. Angle A is then the last angle and would be 180-50-110 so it would be 20 degrees.

[@francescolasaracina3964]

Mike Percival I did it like you did just by watching the thumbnail. I don't know why the author of the video made up all of those lines making a very simple problem look harder than it really is

[@st007tg]

@@Slaave how is F IN EFA 50?

[@sharonminsuk]

Yes, +Mike Percival, how is F in EFA 50? I proceeded exactly as you did up until that point, but I don't see how you get that. Please clarify!

[@sharonminsuk]

Or +Francesco Lasaracina, since you did it "just like he did", maybe you can clarify that?

[@gopakumarkv1216]

FIND THE POSSIBLE ANGLES BY THE RULES THAT LEAVES US WITH THE SUM OF X AND THE ANGLE CFE BEING 110.IF SO TAKE THE BIG TRIANGLE AEB WELL GET

[@michaelrothe5703]

it is really much easier: just call the crossing point of BE & CF D. So CDB is 180-60-50=70. CDE is 180-70=110. The same goes for the mirror/opposite triangles. CAB is180-60-50-30=40. CFB is 180-110-20=50. AFC is 180-50=130. AEF is 180-40-x.CFE is 180-50-30-x. AEB is 180-40-20=120.

[@danah.1578]

why do I get 40 and not 30...? 🤔

[@cannonballbob6949]

In the last result the line is between f and g not c and f, I also got this confused 😂

[@playerunknown7211]

U r doing it wrong Simple

[@koushikmurthy677]

It is 40 only

[@koushikmurthy677]

He is wrong

[@bakashiro]

@@koushikmurthy677 it is 30

[@Whatever-he5gf]

Just have doubt. In 5:57 , in the triangle BFG angleB+angleF+angleG=180 angleG=70 which shows that BG is not equal to GF I just think that this question is a typo

[@MisterJingo93]

I tried to do it with equations and that somehow doesn´t work because they always cancel each other out and I ended up with something like 180 = 180. The solution is right, because the 70 degrees are between the FG and FE line not FC and FE, therefore the middle angle isnt between BE/FC but between BE/FG, which is not the 70 degree angle that can be calculated without the extra lines.

[@rcnayak_58]

In fact, it is not as notoriously difficult as thought. A little effort can solve this geometry problem quite comfortably even without holding this 20 degree key concept! Here it is. Just turn aside the same triangle ABC to the other side of AB (keeping line AB as fixed), making the new triangle ABC'. The point E on line AC, say now becomes E' on the line AC'. Now angle ABE' = angle ABE = 20 degree. Our required angle BEF now becomes the angle BE'F in the new triangle ABC'. But in the triangle CE'B , angle CE'B = 30 degree since angle BCE' = 50, CBE' = 60+20+20 = 100 degree. Therefore, angle BEF is also 30 degree (proved). Draw the triangle and see how quickly it is proved!

[@secretunknown2782]

Mind your Decisions : Draws a 20 degree angle using protector Me : then why the heck you did not use protector to measure **x**

[@Xnyu-dw3uj]

protector

[@secretunknown2782]

I typed protector

[@spacebearcadet746]

Because the diagram isn't always up to scale, for people just want to watch the world burn.

[@secretunknown2782]

@@spacebearcadet746 protectors are not allowed in exams

[@anandk9220]

@@secretunknown2782 Just to politely inform you- It's called protractor, not protector.

[@mint301]

I didn't add any triangles and worked it out. I got every angle of the original shape worked out except CFE, AFE, AEF and x, then trial + error-ed numbers for x using excel in combination with my knowledge of AEB and CFA until x worked. Kind of a trail and error simultaneous equation thingy, I guess I'd call it? I think my way is better as it gets the angle of every single triangle in the shape, doesn't involve loads of sketchy-lines, and it only took my trying 2 numbers for x before I got it right! ;P

[@cecesoclean4591]

Not sure this is right, the sum of the interior angles in a trapezoid is 360°, but in this problem he has the given and solved angles as 80°+80°+120°+70°=350°

[@anilkumarsharma1205]

by using exterior angle we solved it quickly

[@sseb534]

How ? Can you please tell us?

[@prankie727]

I would love to see that

[@notanenglishperson9865]

Ok, I have a Solution. CEB = 180-60-(50+30) = 40 CEB = 2EBF, meaning CBE = 2BEF => CBE = 2x 60 = 2x | /2 x = 30 You might think "it doesn't make any sense" I'll answer, yes, but actually no

[@notanenglishperson9865]

Yo, this doesn't make sense at all, you just randomly guessed it

[@martinbarta7805]

Step 1) solve the third angle of The triangle BCY (the point where Be and CF intersect is Y)=70 degrees 2)if Y is 70 degrees then the angle opposite of Y is also 70 3)we have the first angle of EFY 4)since Y is 70 then 70+__=180 __ is 110 degrees. So we have a triangle with 20,110 and ? Degrees. ? Is 50degrees. 5) so we have another triangle consisting of 20,110 and 50 degrees. 6)looking at F we have 3 different areas next to it. Two of which have the same angle,we know one is 50 so the other one must also be 50,leaving one angle left that must be 80(50+50+___=180) 7)and we are nearly done,we have 80,70 and our final X degrees,80+70 is 150 so X must be 30.

[@codekillerz5392]

Here's how I did it: Solve for the missing angle of CBF by 180 - (50 + 60) = 70 deg 180 - (70 + 20) = 90 Opposite angle must also be 90 180 - (90 + 30) = 60 Assuming angle of E is 90 deg, 180 - (60 + 90) = 30 deg

[@awxangel6781]

CBF has angles 50, 80 and the missing 70. The 60 you're reading from the diagram is CBE.

[@triexortism]

I got the answer 40. I thought i finally solved my first question :'D

[@vallyhangan1069]

cos x =0.86 (Theorema of cosinus in triangles ABC,ABE,CFA, AFE and finally in triangle BFE) tip : you can consider BC = 1

[@lindanot]

I got 30 but solved it in a different manner: I started by figuring out the angle in the left triangle which is 70° (as 180-60-50=70) then by the rule of opposite angles also the one in the triangle with x has to be 70. Then since all angles on on line have to add up to 180° we can figure out that the other angles in the center must be 110 (180-70=110). Then we notice that both angles at the base of triangle ABC are 80 degrees which implies that the top angle has to be 20°. Then we notice that triangle AEF is similar to the top left triangle with angles 20-110-50 so the angles have to be the same. Therefore since both angles at the side of vertex F are 50°, the angle between them has to be 80. Therefore x must be 180-80-70 = 30. No extra lines required! Someone tell me if I’m wrong!

[@kaiwenli989]

I think the step, "Then we notice that triangle AEF is similar to the top left triangle with angles 20-110-50" has to be elaborated upon, as it seems to be nontrivial. However, the rest seems to be correct.

[@aleksag1622]

I did it in the same way

[@ToXiC-vg3xe]

You can't just assume that they are similar (AEF and top left triangle) so you're wrong? nice try tho.

[@SingNostalgiaWithAmogh]

​@@ToXiC-vg3xe ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️

[@ines1289]

it took me like 5 minutes to solve this, it’s kinda intuitive if you don’t get into it thinking it is gonna be hard

[@siddharthadas4155]

Sorry. Couldn't understand suddenly how did you conclude that BG=GF immediately after drawing it. This is critical to your solution. Please explain.

[@TheMIroslaVbachev]

Yeah exactly..I was hoping to find this in the comments. He didn't explain how he got that equilateral triangle

[@siddharthadas4155]

Miroslav, u r right. He didn't explain it. I guess he should have done it instead of spending so much time in earlier triangles on trivial issues. Let me fill it in. From isosceles properties, we get BC=BG and BC=BF. So BG=BF. Hence triangle BGF is an isosceles triangle where angle BGF = angle BFG while the third angle GBF = 60. So all three angles turn out to be equal making BGF an Equilateral triangle. Rest is simple.

[@timothyjholloway]

You didn't say it was an isosceles triangle until you started giving the solution.

[@timothyjholloway]

But, adding up two corners of the triangle does show it to be an isosceles triangle.

[@philipyao5989]

0:52 40 THAT ANSWER IS 40

[@philipyao5989]

I started with the triangle with 50 and 60 degrees. I thought that other angle has to be 70 degrees. Then I looked at the line CF and since I got 70 degrees on one side, the other has to be 110 degrees. Then solved for line BE. The triangle with the angle x in it had two angles with 70 degrees so I singled out x and got 40. How is this flawed?

[@rainbow9832]

Using trigonometry I also got 40 wtf

[@burneraccount9100]

I solved it by using rule that states all angle in a triangle equal to 180. 2nd rule when a line is dissected by another line, both sides of that line equal 180 degrees also. 3rd rule I used was the cat corner rule, since I was able to figure out the angle to the left of the X was 30 degrees I then completed the mirror angle which is if one side of the line equal 180 degree, the opposite side will too. So I mirror my 150 degree angle on the opposite (bottom) side of the line. And thus by using the cat corner rule I am able to infer that x is equal to 30 degrees to complete the 180 degrees since it is now opposite of a 150 degree line. Easier shown than explained, but you don't have to pull made belief triangles out of your ass to solve it, like shown in the video.

[@burneraccount9100]

When I solve the angle to the left of X to be 30 degrees I was able to infer X plus the missing angle on the right of X equals 150 to sum the entire line to 180. Incase I lost anyone in my explanation as to why I use 150 degrees.

[@roberta5448]

@@burneraccount9100, Funny that you found x=30 since your solution is not correct. Firstly, ∠BEC=40°, not 30. And how is your "mirroring" (whatever you mean with that) better than introducing triangle BFG ?

[@bambuaraujo8853]

5:00 knee deep in the math

[@knucklepunch5435]

What?

[@AnthonyCurley]

BFG 9000

[@berrickfillmore8796]

Perhaps you could demonstrate this possibility.

[@knucklepunch5435]

densch123 I agree with Berrick. Please explain yourself.

[@krishan9739]

such an basic problem just made harder by all this worthless construction just use asp everywhere and chain it down u will find the answer

[@surojd]

Yap worthless constructions

[@D_Creations]

Simply go with triangle , as sum of all angles is 180 we can get all required angles as many as you can

[@takyc7883]

Thanks fresh lakewater!

[@JLvatron]

Fun problem, I tried! I solved x=60. I think your answer is correcter.

[@scod3908]

Same mistake as me, slanted EF the wrong way in sketch

[@JLvatron]

@@scod3908 Actually, x was 30, so the full angle at E is 70. The slant direction is correct. I was hoping it wasn't, as it would be an excuse for our errors, lol!

[@SingNostalgiaWithAmogh]

​@@JLvatron ✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️

[@NezamSC]

I solved in it much easier way, without adding any lines in it.

[@KOBY128]

How?

[@captainplur2081]

Imagine saying that then not actually posting it lmao

[@ZeveonHDDev]

@@captainplur2081 use triangle BAE and use the sin laws.

[@pepqr4679]

same

[@rikiyosa4907]

@@KOBY128 lets say the point where BE and CF intersects is D (just to help labeling it) since EBC is 60 degrees and DCB is 50 degrees, BDC will be 180-60-50, which is 70 degrees. BDE is equal to FDE so FDE=70 degrees, BDF will be 180-70 which is 110 degrees. and BFD will be 180-20-110 which is 50 degrees. BFD = AFE so AFE is also 50 degrees, EFD will be 180-50-50 which is 80 degrees. Now we know that FDE is 70 degrees and EFD is 80 degrees, so x is just 180-70-80 which is 30 degrees.

[@DailyDoseofRandomTV]

A simpler way to solve this, lets say the central intersection point is G. Angle BAC=20° because 180°-(60°+20°)-(50°+30°)=20°, angle BGC=70°(180°-GBC-GCB) so angle BGF and CGE=110° making BFG=50° and CEG=40° through addition of angles at this point we realize triangle AFE is similar to BFG so angle AEF=110° therefore angle x=180°-(40°+110°) or x=30°.

[@Mo-ww4bh]

How did you realize that the both triangles are the same AFE we know just one angle

[@SingNostalgiaWithAmogh]

​@@Mo-ww4bh ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️

[@ssjbevegeta3008]

The answer is incorrect to me because BCEF is a trapezium which means the sum of its interior angles should be equal to 360° but if x is equal to 30° then the sum becomes 350° which is incorrect

[@jameslo2818]

??? Who said BCEF is a trapezium?

[@ssjbevegeta3008]

@@jameslo2818Well , my bad , should have revised the properties of a trapezium but my point still stands as BCEF is a quadrilateral which means that the sum of its interior angles should be equal to 360 so 'x' should be or rather, would be 40°

[@prasantamukherjee6418]

Not easy as it seems

[@akbarhussain402]

Triangles FAC and BF(intersection of BE and FC) are similar, you can compare angles to find that angle CEF + 50 + 50 = 180. Then CEF(80) + x + 70 = 180. X is thus 30.

[@absolutelynotkaylee]

i somehow got it right without doing all these steps... maybe i'm just good at guessing (:

[@michaelmulla6178]

I solved this problem within the first 30seconds, the triangle CEF has a 30° angle and a 90° angle,so we know the angle at F has to have 60°, now i look at the triangle EF with the cutting point of the 2 lines, we know it has the 60° of F and a right angle(90°) so x has to be equal to 30

[@UjwalaKamble71]

Michael Mülla CEF=90 how

[@manishkumar-gz2ud]

Wow !!! I'm impressed!!

[@defeat7774]

I calculated without mistake and without watch this video.👍🏻👍🏻

[@TheScience_Network]

This problem is not so hard to solve. This is the main problem that Presh Talwalkar solves in very complex ways.

[@pgorynski]

Seems like only the first rule is needed to solve this. You have enough information at the start to figure out all the other angles

[@skr84nj1022]

Can you do a different solution where you write in every angle in each small triangle and each 180° straight line? That’s how I did it and got 30° very quickly without adding any extra lines.

[@drmattccotton4106]

Correct! You and I chose the intuitive approach.

[@cjsotta163]

same i get the answer using that approach

[@anandk9220]

@@cjsotta163 Please explain or share your method of solution. By the way, here's my solution... [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊

[@JackShen]

Knowing the sum of the angles for the 'X' in the center is 360, helps break the problem down easier. This was in another one of those damn cram books my dad, used to make me do problems from. And filling in the unknowns from the all angles in a triangle = 180, trick gets you to the value of X

[@anandk9220]

[ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊

[@shreyash462]

I solved it in 20 seconds by writing different equations

[@rewernan]

Mr Shreyash it is not necessary write nothing

[@mayanksingh4243]

sir , can you tell me what were that equations !

[@chadjwatson]

And with some hand waving, suddenly BG = BF (with no explanation offered). But more amazing, triangles now only have 170° (70 + 70 + 30). I get frustrated with his mistakes.

[@roberta5448]

Nope, they add up to 180°. You are probably mixing up angles CFE (80°) and GFE (70°).

[@hydrargyrumm]

I CAN SOLVE THAT IN 3 SECONDS! I SWEAR!!

[@mitccoo]

What bothers me is the way you just created BG out of the blue and assumed CBG = 20. This should be given at the start, you can’t just invent it?!

[@harikeshkumarsingh4968]

Me who solved this while seeing the thumbnail. " Was this hard !"

[@peterpaul9763]

On what basis did you start by drawing a line of 20 degrees...

[@智障-r3e]

I found another proof of it. Let the inner center of triangle BCE be I. Extend line CI to J such that IJ=IE. Because ∠CIE=120°, so IJ=IE=JE. Because ∠JEC=60°+∠IEC=80°, so that △BCE≌△JCE. So IE=JE=BC. Because ∠IEB=20°, so we have △IEB≌△FBE. So x=∠IBE=30°.

[@anandk9220]

Which center of triangle is I? Is it the incenter or circumcenter? And kindly explain your solution after this step.

[@智障-r3e]

@@anandk9220 incenter

[@智障-r3e]

@@anandk9220 J is **taken** on the ray CI such that IJ=IE.

[@anandk9220]

@@智障-r3e Now I actually realized myself what you meant. I realized this now before reading your comment. 😊👍

[@智障-r3e]

@@anandk9220 so can you understand my proof now?

[@gideon_kang1]

huh? I'm not sure how everyone in the comments don't realise that you can figure out the method in a few seconds. you don't have to draw anything at all! this question doesn't need to use ANY isosceles triangle logic at ALL.

[@Achill101]

How did you solve it then?

[@junrandyduran2288]

How did you arrive in assuming that line BG and BE are of equal length? How did you arrive in assuming line GF and GE are also of same length? You didn't show it, you just assume without proof

[@niteshm6]

Good notice.

[@prestonharrison2140]

Look at the angles of triangle BGE, it is iscoceles so the two sides must be the same

[@niteshm6]

@@prestonharrison2140 First he said that 2 sides are equal then he come to the conclusion that it is Isosceles triangle. You are going reverse.

[@indianentertainment3193]

I think you are wrong, because I have got the value of "x" is 40.

[@user-lo9wx9mo4z]

more arguements?

[@lesedingoniwildlyluphahla8377]

Am writing tomorrow 😢, needed this

[@ramsuarez82]

I have seen this problem in the Solid Mensuration book.

[@manjunathshivalingaiah1171]

Let's consider the common vertex of the for triangles as O. BOC will be 70. FOE will be 70 vertex angles. BOF+FOE=180 linear angles. So, BOF and COE= 110. OFB=50, CEO=40, OFE=70 isosceles angle. Now consider BCEF the sum of interior angles is equal 360. Therefore x is equal 30. I missed a point and an explanation I think 🤔 nvm

[@garyxlanurias4574]

How is it I got x=40? I started BCA triangle 60+20+50+30=160, so

[@crimsonplaysgames9056]

If angle FCE is 30 degrees and angle CFE is 70 degrees according to the video, then lets label the center point P. Then, angle FPE would be 80 degrees by the 180 degree rule of triangles. Since, FC and BE are intersecting lines, and opposite angles created by intersecting lines are congruent,so angle FPE = angle BPE, so angle BPE is also 80 degrees, but adding the angles of triangle BCP results in 60 + 50 + 80 = 190 degrees. So wouldnt this make the figure impossible? Or am I thinking too hard?

[@ahmedtawfik1642]

(BFC) = 50 SO THE LAST ANGLE IN THAT TRIANGLE IS 180-(50+20)=110 AND LET THIS POINT BE D THEN THE ANGLE THAT COMPLETE IT IS 180 - 110 = 70 THEN (FDE)= 70 , RIGHT? HE SAID X = 30 AND (EFC)=70 BUT 30 + 70 + 70 = 170 IT'S NOT 180! DID I MISS ANYTHING?

[@Coolbreeze02050]

That is not the correct answer at all. If you look as the 50 degree and 60 degree triangle that tells you that the third angle has to be 70 degrees that way it adds up to 180. That 70 is opposite another angle, making that angle also 70 degrees since making a line 180 would make the adjacent angles 110. If the angle in that triangle is 70, I add your 70 and 30 to that and you get 170 degrees in a triangle, which is not possible. The actual answer is that x is 50 and the other angle in that triangle is 60, which adds up to 110, making the whole triangle 180. You didn’t need to add any extra isosceles triangles in there or do any weird bs to get the correct answer. Use the triangle at the right end to find out that x with its adjacent angle to the right has to equal 140 while the same angles above it have to add up to 130. Also knowing that x and the angle across from it in the triangle have to add up to 110 gets you to the conclusion that x is 50 and the other one is 60.

[@roberta5448]

Wrong. "your 70" probably refers to ∠GFE. ∠CFE is 80°

[@realityobservationalist7290]

So I'm trying to understand something I know I've forgotten in math. If you set up a system of equations to solve for x and the three other unknown angles, you can set up a 4x5 matrix... 4eq's/4unk's. Then if you try to solve using matrix algebra (rref), no solution exists. But if you start with x=30, there clearly is a solution for the system of equations. Why is this happening,?

[@justinmplayz8809]

The quadrilateral angles sum up to 350 degrees which shouldn't be possible so this solution is inaccurate. Though I still think 30 is still the answer except the right part of the intersecting lines is 70 degrees and the top part of the same triangle is 80. (50+30)or C+(60+20)or B+(50+80)or F+(30+40)or E=360 which is the correct angles not the bs 70 degrees on the top part. Please review your work before you teach bs to millions of people

[@philiplauren7024]

Except for the fact that he is wrong. Angle G can’t be 40 degrees, it has to be 30 degrees.

[@lmnefg121]

I have got a linear equation system but it is singular. Using line FG introduces an extra independent equation.

[@AlexTikhonovich]

Ohh, in 2:44 angle CGB equals to 70° instead of 80° (180°-60°-30°-20°=70°), and then going through the video answer is 25°, not 30°

[@toky2590]

There are 50 & 30 angles not 60 & 30

[@1millionshroomz838]

Uhh correct me if im wrong but isnt x = 40? gyazo.com/7d70cc51be174554ffb7ffe9c6ab9753 Picture for easier way to explain. So if we take the triangle BCE (A+(B+C)+F) We know a (60) and b+c (80) then F is 40. We need F to solve this. Now we see that F is linear angle pair with G+X a line is 180 degrees F is 40 (we know this from last equation) Then g+x must be 140 (180-40=140) Now we have what we need to find out ONLY G G = x + 40 - 140 = 100 So now we know F is 40 and G is 100 so due to them being on a line (which is 180 degrees) we just take 180-40-100 = 40 = X X = 40

[@adrianocerrat2269]

6:41 How do we know that?

[@asrashujayat4465]

I don't get why we r striking our head so hard ALL U HAVE TO DO is suppose f =x and E is already x and B=20 then x+x+20=180 while x=30

[@romeosierra4431]

It's wrong, If u mark the cross at centre as O then angle COB=70, so angle FOE=70(vertically opposite) And as per ur solution, angle CFE=OFE=70 So, angle FOE+OFE+FEO=180 70+70+x=180 x=40 Pl comment.

[@XenoWars]

This doesnt really add up. At the start before doing anything, you can see angle(B,the middle,C) is 70°. So from that, you get the two angles to its left and right are 110°, which in turn makes the remaining angle in the center 70°. But if angle(C,F,E) was 70° that means x would be 40°.

[@DistinctDoge]

Thats what I did and i also got 40, i wonder if there is something wrong about the question or the way we did the math

[@StudyTutor91]

Hello dosto . Mera youtube channel bhi dekhlo please agar pasand aye to like 👍 and subscribe Dosto mera youtube channel mai aapko class 9 ki physics ,maths ,history videos milenge .Please but atleast dekh to lo.

[@Intrafacial86]

This is what I hate about math problems. If you give up and look at the answer, you essentially poison the mind and can’t make yourself forget to be able to figure it out in your own way. And yeah, just “knowing” which isosceles triangle you needed seems kinda cheaty. I feel like there’s a proper way to draw back from “x” to a point where you can get a triangle to start with

[@Deus69xxx1]

base angles in the diagram show the answer to be incorrect. unless i'm way off base. read on! if i'm allowed to measure out an isoscelese triangle, why am i not allowed to use said tool to measure the angle inquestion? if you have the tools to guarantee you're making a proper angle, you have to tools to just measure the answer. doing things the way described is the same as using a calculator on your multi-line multiplication problem. as far as i can see, and my mathing out all the angles of all the triangles, this entirely hinges (without a tool) on assumptions. in which case, i'd guess in a range of 20-40. additionally, when adding up ALL the triangles you can figure out, the unnamed angle in EF? is 70 degrees, which is counter to the answer of point F in EF? being 70 if the answer is 30. the angle math makes that center crossing two sides at 110, two at 70. (both B and C are 80, as the two angles in those corners add to 80. that leaves A as 20. CFA is 30-?-20, which means point F in CFA is 130. BEA 20-?-20, which leaves point E as 140. consider the center cross to be labeled G. creating a square(lol) of GFAE, you have 140+130+20+? = 360. that would leave GFE's G spot to be 70. from the other side, you have CBG at 50-60-?, and that makes the G spot 70 on the other side. since that's two even angles, you can now make a square of the center, with two 70's and two 110's. both of these proofs of the self labeled G spot counter indicate that x can be 30, if the angle you found dictate that the F in GFE is 70, and it's proven that G in GFE is 70, there's a problem.) UNLESS i failed somehow, which i'd like to know where =/

[@duncansmol9362]

Damn, you really made this unnecessarily difficult.

[@x-x2060]

True he did make it more difficult, theres other easier ways to figure it out

[@atikshagarwal5147]

What is the easiest way?

[@x-x2060]

@@atikshagarwal5147 FC and BE make an X and when theres an x theres 2 same angles that are opposite to each other.

[@davefickess7973]

@@x-x2060 Using that and exterior angle theorem is how I got it.

[@xxlguitarscom]

@@davefickess7973 could you elaborate please?

[@hessylaguna5415]

solved it! Before watching: this is how its done; look at Triangle BAE. this triangle is in a relationship with a different triangle so it spells "bae". We must find this triangle's bae by using the boyfriend line (line BF). We see that triangle BFE and BAE share a common fetish known as "x" this is reasonable proof as to why they are together. But what is their fetish? That is what the question is asking. lets look at other clues. notice how there is no D in the picture. That is because the D is in the middle (lol). The intersecting piece must be "d" and the angle which contains the fetish is named FED. This is not a coincidence, they have a food fetish which only about 30% of triangles have! if you were to put this into math talk then "x = 30 degrees" and that is how you solve this one.

[@total_dk6517]

Nice one. 10/10

[@androidkenobi]

TEEHEE!

[@llavishxd]

I thought you were serious until the end 😩

[@danziv94]

I knew it its like my mother always said u gotta look for that D.

[@keenantroll5151]

did you take into account the GBF, gay best friend triangle?

[@spiderjerusalem4009]

Hillarious that people here have been complaining about how the OP's attempt made it hard, but when they try to elaborate it 𝗴𝗲𝗼𝗺𝗲𝘁𝗿𝗶𝗰𝗮𝗹𝗹𝘆, they either somehow suddenly know the measure of one of the unsolvable angles or merely straight ahead solve it trigonometrically The problem with the "oNLy cOuNtiNg, nO dRaWiNg" method is that you would always end up getting "both-sides-equal-equation" such as 180°=180°, x=x, etc, as the final count and result. You'd always get stuck on that. You wouldn't get anywhere or even move forward. if we merely had to know that sum of all triangle's angles equal to 180° and square's equal to 360° and vertical angles equal to each other, then such question wouldn't have come out in the first place. I've doubted these comments for a long time. They either solved it trigonometrically without having read the title or commented another "yOu pOinTlEsSly mAdE iT hARd" merely for attention and reactions due to what the majority of the comment section have been saying

[@anandk9220]

[ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊

[@necrolord1920]

nice problem IMO I tried solving it as 4 equations and 4 unknowns, but the equations I used were no good

[@chlorophyllphile]

me too. there are dependent so no way to solve :(

[@Peter_1986]

I tried that as well and I got a solution like x = x or something, which is obviously a completely meaningless answer.

[@captainobvious9581]

+Laurelindo well, you are right, the value of x is equal to the value of x

[@lordwalterebenyzrbautista204]

Yeah, but it's still meaningless when trying to find th- *realizes you are Captain Obvious* Oh....

[@KAF128]

I did the same !

[@TheOoorrrr]

I see people keep saying "I solved it in a different way, it's an easy question"... Ok, so share the proof instead of bragging about it

[@Skimbleshanks73]

I tried it aswell, it is actually possible, by just adding up the angles you're missing 2 angles that can be replaced by y/Z then you can form an equation with both of these and solve for Z and y then you have all the angles and can get to x, I wrote that down somewhere earlier don't want to do it again tho

[@demifiend9]

@@Skimbleshanks73 It's not possible. There's not enough information to solve those equations. Go on and actually try solving those equations and you'll understand.

[@tsunholiu4441]

@@demifiend9 It's actually possible by applying sine law, but the calculation is a little bit messy and compound angle formula is needed for solving the trigonometric equation. The method in the video is much more beautiful

[@kevinmartin7760]

@@Skimbleshanks73 Yes, you can generate two equations in two unknowns but the equations do not produce a unique solution. No matter how you try solving the two equations you only end up with platitudes like 0 = 0 or y = y. If you try expressing your set of linear equations in matrix/vector notation, you will find that the matrix has a determinant of zero and so cannot be inverted.

[@sandeepbannore4185]

@@demifiend9 x I think can have many values. I might be wrong as well

[@danielbuttons5535]

What If you used triangle between angle 50 and 60 degrees then that angle will be 70 degrees and since opposite angles are equal the other angle is also 70 degrees.

[@tabibgd7880]

daniel buttons I did the same thing :P

[@juanmatias87]

If you add up the opposite angle by this method and the results they don't add 180 degrees

[@modellocomotiveworks2215]

@@juanmatias87 opposite angles don't provide 180 degree. Linear pair angles ( angles on a straight line) gives 180 degree and that's right in this question.

[@loveculture5250]

Yes me too and my x was 40 degrees because of that.

[@vicenteizquierdo5063]

As vpn x is 60

[@entubatumahumasu4132]

5:04 you mean BG and BF

[@mulimotola44]

exactly. He should fix it in an annotation

[@MindYourDecisions]

You're right, thanks for letting me know. Sorry for the mistake, I have added an annotation and included the correction in the video description.

[@twanwilting3770]

He ment BigFriendlyGiant BFG

[@santiagoarce5672]

Presh, I think you got x wrong. This is unless crossing lines don't create equal angles on opposite sides. Maybe I am wrong, but, in the triangle with angles 50º and 60º, the remaining angle is 70º, and so is the one on the opposite side . But the problem with your solution comes when you add this angle to F, which is 70º and x which is 30º. It doesn't add up to 180. it adds up to 170º

[@Hbaterinha]

Almost. The angle you found is related to the line "C-F". In the final drawing, when he finds 70º, the line he uses is "F-G".

[@hookerWithATool]

I solved it using parallel segments you can draw a segment which is parallel to EB from A, then you can connect the segment to point F which would give you two congruent triangles by SAS. The rest is pretty simple actually.