A Classically Hard Geometry Problem

Some call this the "hardest easy geometry problem." It seems easy but it is incredibly hard to solve using elementary methods (no trigonometry allowed).
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[@yourfriendlymintlemonade]

Only the true fans of Presh Talwalkar will notice this 'r' upload !

[@atharvius7522]

What? Where??

[@Shreyas_Jaiswal]

I know that but it's a little different question Check his old video kzbin.info/www/bejne/foLGXmhqnbZrodk

[@henrilaporte7599]

This is because he wants to collect more money from the same video.

[@warfyaa6143]

Only the true fans of Presh Talwalkar will notice this is not a re-upload of a previous similar problem which has 50-30 angles at B not 60-20.

[@MurtadhaAljanabi]

@@warfyaa6143 And how does that change anything apart from calculations?

[@icarusash2241]

He said pause if u want to try,and I skipped 10secs into, that's my confidence in solving 😂

[@karsoonyong4287]

It took me a while to get your joke haha

[@swaroop4una]

Lamo🤣🤣🤣

[@daviesfamily9608]

lol

[@xxx_arandomguy_xxx6727]

Anyone want 200th like? (I got 199th)

[@10thTime-gi8qt]

I solved in in 3 days "just 3 days..😅" but I swer I didn't see the solution, Now I have it's easiest solution from which it can be solved in just 2 mins. ( If you didn't believe me reply me will give you it's solution, you will not find that easiest solution any where)...{ I am an Indian 9th standard student}.

[@aadhya8459]

I always know that something's to be constructed when geometrical problem seems to be hard. But the problem is: Either I just don't construct it(thinking, how smart I'm and I'll be okay with this much) or I end up with getting confused what's needed and what should be there.

[@nandisaand5287]

I mentally broke the figure up into different triangles to determine all the angles. Got all but 3. Then I wrote a bunch of eqns hoping algebra could get me there (10+X+C+20=180; 20+X+C==180, etc), but none of that worked. It didn't occur to me to make even MORE triangles!

[@twwc960]

A lot of people are claiming that he solved this problem before in a previous video (The Langley Adventitious Angle problem), but with slightly different numbers. The fascinating thing about this problem is those innocent looking changes in the given angles change the problem completely! Try using the solution he gave in the earlier video on this problem: it won't work. That solution was very specific to the angles given, allowing the construction of an isosceles which in this problem would not be isosceles. A completely different solution is required. They are both fiendishly difficult problems!

[@Achill101]

But what's the point of being limited to elementary geometry for a fiendishly difficult problem when the same simple trigonometric solution can be applied to ALL problems of that type? Isn't that a sign that trigonometry is great? I used my trigonometric approach to the Adventitious Angle problem again and solved this problem in a few straightforward lines: x is 20 degrees. (Calculation in my reply to myself.)

[@Achill101]

We can use the sum of angles in a triangle to get angles and a first equation for x. ADB=40°, AEB=30°, x+BDE=130° We get a second equation for x through the sine values for the corner angles of the area ABDE. For each corner, we can define a left angle and a right angle: for corner A, the left angle is EAD and the right angle is BAE. Then the product of the sine of the four left angles is equal to the product of the sine of the four right angles: sin(BDE)*sin30°*sin60°*sin10° =sin(x)*sin20°*sin70°*sin40° The two equations lead to tan(x)=0.36397 or x=20° (using sin(130°-x)=sin130°*cos(x)-cos130°*sin(x))

[@lamcho00]

I think this is one of those tasks for elementary students, where they are testing if you can construct a triangle to scale and measure the angle to find the answer. I don't think they had to mathematically prove the solution. Though it's an interesting solution and I couldn't prove the solution on my own either. I guess finding what to construct requires weeks of trying different constructs or a lot of experience solving similar problems or just plain luck.

[@RahimmacDonald]

I always found geometry to be harder than algebra

[@richardleveson6467]

Very nice problem - real old fashioned geometry! Everything depends on a shrewd construction - thank you!

[@coachhannah2403]

Solved it without adding any lines. Fill in all triangularly determined angle. Angle c = 20 < BDC = 140 < AEC = 150 < ADE = 40 + y < BED = 30 + x From tri CDE we get x - y = 10 From quad ADEB we get x + y = 130 (i.e., 360 - 80 - 80 - 40 - 30) Algebra gives y = 60; therefore x = 70

[@Achill101]

x+y=130° is correct. But could you explain how you get x-y=10° from triangle CDE? I don't think it is correct, instead I get x-y=-90° from my (trigonometric) solution for x.

[@coachhannah2403]

@@Achill101 - Well, that WAS 3 years ago, so, either I made a mistake, which is totally believable, or I'd have to reinvent my steps, which is unlikely.

[@Achill101]

@coachhannah2403 - only work on it if it's fun. Happy New Year.

[@Windmill97]

Me thinking I was so close to solving it watches solution: “I was nowhere near solving it.”

[@抬大樹學歌]

I kept thinking this problem until I fell asleep last night. But I finally solved it this morning!

[@Leflaneur-412]

well i did try to solve it , but then i realized that there aint enough information to calculate the angle , so new lines must be constructed , therefore i didnt carry forward. But thats a fun problem tho , tho i didnt committed to solve it , since it is somehow quite energy-consuming LMAO

[@JohnDlugosz]

I also thought it needed more lines, but then (after labeling every angle) I spotted a pair of similar triangles that were identical triangles. That allowed me to add another, independent, function in the same unknowns.

[@cluelessarn]

@@JohnDlugosz which two may i ask.

[@astik54]

yeah, me too, but even after i constructed a few lines, it wasn't enough to solve the problem

[@Leflaneur-412]

@@astik54 well let me provide some insights , but still tho trigonometry is needed , BD=DC and AC=BC , let one of the sides be x, then sides like DE AE and AD could be found , then sine or cosine theorem is enough to solve it.

[@Leflaneur-412]

but well i mean trigonometry is deeply related to geometry, so i dont regard that as cheating somehow LMAO(since the problem requires us to solve with elementary geometry only)

[@Sinister_713]

Damn it was awesome. Keep making these videos. It helps me a lot!

[@FALLEN_GOD]

It was surprising easy for me. I just had to find out all angles in my head.

[@mustafaberent7785]

Outer lines makes it easier than stucking in the triangle

[@francisadriangulmatico1356]

using trigonometry laws, you can solve it in a glance which is 20°. However, without using these, it's really hard Lol

[@A_Name_]

When I saw it I thought the same thing.

[@kiranbarnwal8503]

How can we use trigonometry?

[@Secret64462]

I'm a middle schooler and it took me 10 minutes of just looking at the thumbnail to get x=20 No wonder they said use elementary methods

[@joeyhardin5903]

ok heres how i solved it in 1 minute (no equations or unnecessary line construction:) first label every angle you can on the diagram, using opposite angle theorem, straight line 180 degrees, triangle angle sums, etc. then construct a line from D parallel to AB and mark F as the point where that line crosses BC. use the corresponding angles theorem to show that BDF = 60 degrees. mark G as the intersection between AE and BD. note that angle DGE = 50 degrees which hopefully you worked out earlier. mark H as the intersection point between AE and DF. consider the triangle DGH. its angles are 60, 50, and 70 (we already knew 50 and 60,) hence angle FHE = 70 degrees also. from here consider the triangle HEF and notice that angle HFE is 80 degrees. now look at triangle DEF and calculate angle EDF to be 50 degrees. finally consider the triangle DGE with sides 110, 50, and the missing angle, 20 degrees.

[@amlan1098]

How u got angle EDF to be 50° at last : ?

[@joeyhardin5903]

@@amlan1098 i cant remember, but i mentioned at the start to label every angle you see. they can all be calculated with middle-school level angle theorems

[@tuulenkhsuren380]

I try to solve this myself. then 2hours later finally I solve this. it was soo hard

[@felipegaspar657]

Genius solution!

[@khangkhuimakang5012]

Y is the biggest triangle's angles more than 180.

[@lifechangertutorial]

I'm amazed at how u solve your problems. I hope to do same in the future.

[@gaurei-02j70]

Thank you very much sir and keep making more🤩

[@DHaLMediaEnglishOfficial]

OMG! I was wondering if he is going to use all the letters from A to Z for extra construction. 🤔 Loved it anyway.

[@qiu_shui]

My method is to list equation using the law of sines and then solve it by Mathematica.

[@sergioavalos7399]

I remember my geometry teacher of the fourth year of high school gave us this problem and got upset cause anyone couldn't solve it

[@sukursukur3617]

Thanks solidworks. It took 15sec

[@noorahmed6662]

2:58 the triangle is impossible to be formed! I think this solution is not completely correct! 😅

[@johnbatchler8551]

You have solve this problem using vertex method why u been given two angles and u need to solve the third angle another rule to use verticle angles are congruent

[@idkwtvr4844]

I drew all the correct lines, but failed to see the big isosceles triangle.

[@jerry2357]

Define the intersection between lines AE and BD to be H. From Presh’s solution I notice that triangle BHE is similar to CDE, which can’t be coincidence. Also, because angle DBC and angle BCD are both 20 degrees then triangle BCD must be isosceles, so BD = DC. I’m just wondering if there is another way of tackling this by proving that BHE and CDE are similar, although I can’t see a way of doing this at the moment?

[@AA-vy9dv]

For more generalized angle problems like this, search “Langley’s adventitious angles” on wikipedia.

[@prasenjitpoddar1839]

How do you know that CG will be bisector of ACB?

[@MonsieurBiga]

2'20 : "notice that this triangle is exactly congruent to this triangle". Why is that? What if the segment DF was pushed up for example?

[@liyi-hua2111]

Monsieur Biganon because in this case we have ABC congruent to CDF and they are isometric

[@cucgiongphangiang2955]

why fd=fe?

[@Uddaybhaikabaap]

Presh u've already posted this same earlier earlier I guess When u said it's the most easy looking evil question

[@chaitanyasoni6144]

You have already given this type of question in another video of yours

[@RayoIII]

I thought this question was already solved in this channel? I had a similar question (in highschool) but just with some X-es and find what is X.

[@deenislam2965]

I solved this at 2012 by a different method.

[@witkotamm7480]

Isn't this an reupload from a few years ago?

[@leopoldhoang3103]

Can't it also be 30 degrees?

[@RicoGrafity]

its took anhour and 10sheet of paper for me to solve this

[@gundeepsaini8756]

I used coordinate geometry. Pretty simple after that. I hope I didn't cheat!

[@معاذالدوسري-ث6م]

Solved it using fixed point point method and checked for all the angles which turned out to be correct x=20*

[@sntripathi84]

I hate geometry but I loved the way he had solved

[@KsharkBlue]

After I saw this problem, I thought it's very easy. But After I try, I thought this video is not unmeaningful. I try and try but it was same. +x-x, +x-x ...

[@wyattb3138]

The problem was that there were no parallel lines.

[@iuliusconstantcornelio2018]

A triangle question WITHOUT GoU gU ? What a blessing....

[@FootLettuce]

GouGU is trig

[@johnjordan3552]

I personally prefer Pythagoras Theorem

[@sayanlaha3649]

​@@johnjordan3552no, it's Brahmagupta's!

[@xz1891]

This is the famous "Hardest easiest geometry problem". I remember it's 20° or 30°.

[@user-xx6in8vu6c]

No it was a variant, it starts off the same way but changes later, presh has done a video on it in the recent past

[@rgabor90]

This is a brilliant solution, I was thinking like 1 hour, when you posted your tweet, but I couldn't come up with a solution, but this was beautiful, it made my day, thank you! :)

[@KodamatiUdayaPrabhuKiran]

When I saw it, I thought I can solve it in minutes and opened paper and started. 5 minutes later! I am like, this is why it is quoted as "Insanely Hard."

[@MindYourDecisions]

I admit I didn't solve it. But I'm sure the people who did solve it will at least understand that others find it difficult. (Since people always ask, the comment is from before the video is published due to advance preview).

[@Dreamprism]

I felt kind of the same. After a bit of not seeing it, I was like "well I know I could do this with a coordinate system and trigonometry if I needed to".

[@leif1075]

@@MindYourDecisions Don't you think everyone would find this difficult even those who solved it?

[@bhabanishankarrath3973]

@@MindYourDecisions This is not at all hard. In india my seniors i.e class 10th's student used to do this in like seconds. I am in 7th and acoording to our cpncept a triangle's whole degree's sum should be 180 degree and then as follows - I solved it in this way tbh o knows that this is wrong but i got correct answer. Its something like that--- As we all knows that an triangle have 180 degree at all so what i did is 70+10+60+20=160 and 180-160=20 degree. this is how i did this without even using algebra. Even that comment is from an indian uday is a indian name.

[@ayan8233]

@@bhabanishankarrath3973 it's incomplete lol

[@yasharya9765]

I solved it. Now I can sleep better 😴

[@adeeb1787]

Yes, me too.

[@adeeb1787]

Weekends are good for sleep.

[@yasharya9765]

Yup 😍

[@cluelessarn]

Did you use the same method ??

[@yasharya9765]

@@cluelessarn yes the approach of the problem was to observe the isosceles and equilateral triangle using construction.

[@i_windsurferyeah346]

i've solved it in another way, in case of the prohibition/banning of drawing additional lines. in triangle ABC angle ACB =180º-80º-80º=20º. line AE and line BD cross in point O. in triangle AOB angle AOB =180º-70º-60º=50º -> in triangle BOE angle BOE =180º-50º=130º and angle OEB = 180º-130º-20º=30º. in triangle AOD angle ADO =180º-130º-10º=40º. Now, it will be quite complicated: x = angle OED , y = angle ODE , a = angle EDC, b = angle DEC. triangle ODE: x+y+50º=180º -> x+y=130º triangle CDE: a+b+20º=180º -> a+b=160º line ADC: a+y+40º=180º -> a+y=140º line BEC: x+b+30º=180º -> x+b=150º ,so we have a system. With y=140º-a and y=130º-x, we have 140º-a=130º-x -> a=x+10º. With x=150º-b and x=130º-y, we have 150º-b=130º-y -> y=b-20º After solving this system, we'll get x=20º, y=110º, a=30º, b=130º.

[@Theo0x89]

_"After solving this system, we'll get x=20º, y=110º, a=30º, b=130º."_ No, we won't. You have four equations for four variables, but the equations are not independent: (x+y−130°)+(a+b−160°)=(a+y−140º)+ (x+b−150º)

[@i_windsurferyeah346]

@@Theo0x89 Algorithm was rewritten from paper piece abd somehow i got the 20:-)

[@souradootbiswas905]

Look easy but isn't

[@raviphone5308]

Hi, could you post solving the same problem using Trigonometry. Let's understand the power of Trigonometry. Thank You.

[@areeb6703]

Me: *sees thumbnail and tries to solve* 5 mins later: yea boi x should be 60 degrees My confidence and self esteem after watching the complicated solution and x=20 : I’ve decided I want to die

[@bhabanishankarrath3973]

I solved it in this way tbh o knows that this is wrong but i got correct answer. Its something like that--- As we all knows that an triangle have 180 degree at all so what i did is 70+10+60+20=160 and 180-160=20 degree. this is how i did this without even using algebra.

[@asasadcdbvcvb4334]

i bet you used Trigonometry this why you got 60

[@denizkaragullu6239]

I got 60 as well without trigonometry due to my awful drawing skills

[@areeb6703]

Deniz Karagüllü same..and I drew the figure really small so it all became messy

[@Xn10001]

I got 60 as well using a system of equations, and it's not wrong. X can be both 60 and 20 and still satisfy all the constraints

[@RobW612]

Draw BD so it is 60°? First of all, BD would be a line segment, which is not measured in degrees. Degrees are measured by angles. Even if BD was 60°, what line segment is it 60° from? The answer to that would most likely change the result of the drawing, not to mention a completely different mathematical calculation for all other angles affected. Clarity is the key in Mathematics, not assumptions.

[@universaltravellershostel3402]

I think the last part was explained really complicated. If it's an icoselise triangle with apex = 80°, then you know the other two equal angles must add up to 100° to fulfill the 180° internal angles necessary to form a triangle, so just divide 100° by 2. And you get 50° each angle, then subtract the complementary angle of angle X in FED from 50° to get X°.

[@klaxyrine979]

I was confused about that explanation too. It would be better if it was explained in such way that given the angle of 80 degrees, and another angle of 30 + x degrees, and an unknown angle - just assign 30 + x degrees on the unknown angle then solve for x which should total 180 degrees.

[@Night-o-trix]

I know u just did the everything is equal to everything hence it's 20°

[@VibingMath]

The Langley's Adventitious Angles! Thank you for solving this hardest elementary problem!

[@APRENDEREVOLUIR]

Parabéns pelo excelente trabalho! Ganhou mais um inscrito! 👏👏👏👏

[@jonathasdavid9902]

This is the famous Russian triangle problems, I met this problem with this name. Thanks from Brazil.

[@ANILKUMAR-on6rz]

Hello..

[@raileite5994]

É nós consagrado

[@bhabanishankarrath3973]

This is not at all hard. In india my seniors i.e class 10th's student used to do this in like seconds. I am in 7th and acoording to our cpncept a triangle's whole degree's sum should be 180 degree and then as follows - I solved it in this way tbh o knows that this is wrong but i got correct answer. Its something like that--- As we all knows that an triangle have 180 degree at all so what i did is 70+10+60+20=160 and 180-160=20 degree. this is how i did this without even using algebra.

[@guilhermesalmi5115]

salve

[@ushajain5031]

3:05 I thought sum of of angles of a triangle measure 180 degree

[@policjato]

Yes you have right, he forgot in 2:54 add 60+80=140 the biggest angle, and two other have 20 degree (Tak masz rację zapomniał z 2:54 dodać 60+80=140 największy kąt, a dwa pozostałe po 20.)

[@A_Name_]

@@policjato if you can speak english why did you reply in (dont want to offend anyone) what ever language that is?

[@MidnightTheOne]

Good point haha

[@ushajain5031]

@@policjato haiai8e ufjfndn u82919 hindshhsi jjdjie jdiwowo hsk8qyhe hdjnow rqgajldlbue uidiejbs ikkooaw jaisijdj I hope u understand

[@policjato]

@@A_Name_ Ups sorry men, English is not my first language. This is polish language. I write comment fast, not thinking about it, this time I look English movie and should write in English. My comment:"Yes you have right, he forgot in 2:54 add 60+80=140 the biggest angle, and two other have 20 degree"

[@Perokoulos]

Had to rewatch it after 3 years... I stumbled upon this problem again today but I didn't remember the solution that is indeed complex.

[@10thTime-gi8qt]

I solved in in 3 days "just 3 days..😅" but I swer I didn't see the solution, Now I have it's easiest solution from which it can be solved in just 2 mins. ( If you didn't believe me reply me will give you it's solution, you will not find that easiest solution any where)...{ I am an Indian 9th standard student}.

[@Yochillybro]

Share solution

[@sharingheart13]

I took math all the way up to Calculus II in college, which included Trigonometry, and at no point did any of the problems require for us to draw additional lines into the diagrams given to effectively create a new diagram, so I wouldn't consider this an "elementary method".

[@Achill101]

The solution is elementary in the sense that it doesn't use trigonometric functions like sine and cosine. But the solution is not simple. On the other hand, with sine and cosine, the solution becomes relatively simple.

[@dakshsingh6865]

Triangle CDE is Similar to Triangle BE(intersection pt) HENCE angle DEC=130 and x=20

[@hanswust6972]

Really hard to figure out a solution, happy you share it.

[@2hmed890]

Hello, I graduated from high school a month ago. I really need to improve myself at mathematics especially at algebra, trigonometry and geometry, so I would appreciate that if anyone suggested good books for me💜.

[@HeLithium]

don't forget about calculus ;)احمد خلف

[@burner887]

Get a brilliant subscription (or at least the trial)

[@yourfriendlymintlemonade]

" Methode x " is one of the best .. helped me through my engineering study in college .. i always had another method to solve problems even if the prof doesn't know it haha .

[@Chasval]

For Geometry, just go to the foundations with Euclid's Elements. Tough to read here and there, you will have to google some stuff. But a fun book. It was the first book I bought in my math journey. The second was Shilov's Real and Complex Analysis, which is a difficult book, but made me appreciate the foundations of calculus. I don't actually recommend it as your second book. Spivak's Calculus would probably be where I would self learn calculus, if I could do it over again. But remember to get to it at some point. The best recommendation is just go to the library (a good one attached to a university) and go to the math section. Find short and easy books to play with and figure out what style you are looking for.

[@dhruvjain2388]

You can use books like challenging problems in geometry and challenging problems in algebra. They are very good for both advance and basics. Pdf is available over Internet.

[@hasibulislamshanto1190]

Nah 😔 couldn't solve this using elementary method. Solved it with trigonometry and co ordinate geometry

[@Achill101]

So did I, using trigonometry. How did you do it?

[@peterbentleyhk]

Just for your info Presh : this is VERY old puzzle !* I first came across it when I was at junior school (pre- 12 YO) , over 60 years ago, and of course none of us in the class could solve it at that time until our math master eventually told us that we needed to make a construction (and even then we needed a lot more hints ) I have since seen three different methods to solve the puzzle, all using different constructions, of which your's is one. _________________________________________________________________ *PS: or am I thinking of the "Langley's Adventitious Angles" puzzle which you posted some time ago ?

[@reldahr01]

I wrote down all the equations and started substituting and ended up with x = x

[@sh3n3ng]

and it is technically right.

[@shuvamkumar7978]

After watching the thumbnail for half an hour... i finally opened the video to see the solution..😬😬

[@dovearchambault9229]

I solved it by filling in the rest of the easy to calculate angles, giving the unknown angles variable names, and then attempting to solve for each one until I wound up with -x=190-(150-x), which simplified down to x=-20, which is actually x=20 (Edit: Took me about an hour, so it's not nearly as efficient)

[@jamjam3448]

I now know why it is so difficult. We want more of such questions.

[@bhabanishankarrath3973]

This is not at all hard. In india my seniors i.e class 10th's student used to do this in like seconds. I am in 7th and acoording to our cpncept a triangle's whole degree's sum should be 180 degree and then as follows - I solved it in this way tbh o knows that this is wrong but i got correct answer. Its something like that--- As we all knows that an triangle have 180 degree at all so what i did is 70+10+60+20=160 and 180-160=20 degree. this is how i did this without even using algebra.

[@kimba381]

@@bhabanishankarrath3973 You have determined angle ACB. You get no points for answering a different question to that asked.

[@franciscoparedes1131]

@@bhabanishankarrath3973 Are you ok?? You've just found angle ACB. The question is for angle AED.

[@prathvirajchauhan62]

@@bhabanishankarrath3973 I can't tell if you are serious or trolling

[@mohitrawat5225]

@@prathvirajchauhan62 He is a fool who just solved the question in a fluke. In India we don't solve such hard questions.

[@atomed1]

It is not simple but I have a straight forward solution just by utilizing a simple supplementary angles rule. If you extend a segment from "C" , up and then extend B-D to intersect the end of that segment, it will create "DCF" - 70,70, 40 isosceles triangle. At this point all you have to do is solve for all the angles leading to "X" by either adding to 180 degrees (supplementary angles rule) and/or (sum of angles in a triangle). :)

[@ameerunbegum7525]

Honestly I didnt even think about that.

[@thepronoob1972]

Thanks Presh for solving this. It helped a lot- Umang

[@barrywein2185]

It took me three @#$52 years to solve it pre-internet but, I never gave up and eventually did it.

[@leighmatheson4797]

Use a little intuition. Call the intersection of BD and AE point P. A quick look will tell you that angle DCE equals angle PBE which is 20 degrees and further that triangles PBE and DEC appear similar. To be similar means angle DEC must be 130 degrees giving a value of 20 degrees for x. This value works fine for all possible triangles so the answer is 20 degrees. Of course the triangles might not have been similar which would have required a more rigorous approach but why not first try what appears to be obvious?

[@chitailun]

This problem told us how weak our geometry is. Can you write out a function describing the relationship between x and

[@fishloh97]

Hello Presh fr Singapore. I've a question that's unrelated to Math... Hope you don't mind :). May i know what are the tools, software etc you used to create these great, easy-to-understand visual explanations?

[@anjanasinha5518]

Congruency had left the chat @ rip

🙏🙏🙏please get rid of crappy background soundtrack jingle . You (Math) don't need crud💩💩💩💩💩💩💩💩💩💩💩💩💩💩💩

[@alomirk2812]

It cant make sense imagine point f beetwen A and E and B and D angle AFB = 50 (180 in a triangle) so angle DFE will equal 50 too angle BFE = 120(180 in a line) so angle FEB will equal 40 (180 in a triangle) BDE + DEB + EBD = 180 so 20 + BDE + x + 40 =180 EAD + ADE + DEA = 180 so 10 + 40 + BDE + x = 180 that means 20 + BDE + x + 40 = 10 + 40 + BDE + x we can remove BDE so 20 + x + 40 = 10 + 40 x and that doesent have a solution...

[@kuboskube]

I watched the solution and I still don't comprehend it. I understand the steps you took by identifying triangles, but geez... there's simply too many steps.

[@howareyou4400]

This is actually one of the hardest problems this channel ever touched, especially considering how little pre-knowledge you need to solve this.

[@Achill101]

Then again, if you allow trigonometry, a little pre-knowledge allows an easy solution.

[@Phobero]

Elegant solution; no Gougu = +1 like

[@hemishparmeshwar8454]

I got 70 degrees and it did work out fine. I'm confused now

[@saberamuna104]

x has infinitely many solutions, and the answer is 'x is less than 130 degree'. Check it to verify!

[@alanx4121]

differential geometry, how would x change if 70 was 60 and changed to 70?

[@tsgsjeremy]

Keep getting closer. I try to solve up till I'm sure I'm missing a simple congruency or other relationship, then watch the video. Extending all the lines helps too.

[@iamyoda7917]

Did you do this problem before?

[@UmerQureshi21]

WiseOldDude yes he did

[@hafez591]

I think either it was posted earlier or one that is similar

[@gosoloph]

Same problem but different approach I think.

[@Tim3.14]

Are you thinking of this one? Looks similar, but not the same: kzbin.info/www/bejne/a6fLnJ-IjNSasNE

[@billy.7113]

@@Tim3.14 Not that one.

[@Mathematician6124]

Ah easy problem. Solved it in seconds. With the same construction. But after construction I followed a different method

[@nickcanncann7044]

There is a much simpler method, extend lines at intersections d and e , that gives opposit equally angles, the rest is just isosceles triangle 180 degree missing angles!

[@vishalsawarn4234]

I guess everyone has solved step 1. But step 2 is unsolvable without any hint of construction🧐🧐🧐 After getting step 2,one can easily solve for step 3,,,as I did

[@dawidkalandyk2873]

I solve it and x equals 70... And I can't find where I'm wrong :o

[@GDPlainA]

I found out that 150-x is bigger than 90 degrees for some reason. So, x should be lower than 60 degrees.

[@himanishbasu772]

I admit that as a student from class 8 from India, I had solved the problem using some algebra.

[@gaminginfo1686]

i am still stuck how the triangle ADF congruent to triangle BDF anyone can help me out plzz

[@TheRoacker123]

3:02 isn't angle sum of any triangle 180°?

[@factsverse9957]

Oh AFC = 140°, FDA was not accounted for.

[@johnbatchler8551]

You been given two base angles of one triangle so why don't u add them up to get the third and that is one if vertex angles without comically the problem more

[@cardiosinvestestment]

In your naming of your triangle AFC, do you mean the vertical angle is 80 degrees?If so, then you are wrong! Secondly, the fact that AG=CE,does it necessarily mean GF=EF? I am not satisfied by this mere saying, please make an appeal to a law, axiom , or any other geometric theorem for justification. Thanks

[@cardiosinvestestment]

I think I get it now, only answer the first part of the question , the second part is clear

[@Rhythm12335]

How can we confirm that CG bisects angle ACB?? I am confused.