I would suggest that angle CDE can be directly calculated to be 3x, which is the exterior angle of the triangle EDB = 2x + x similar case for angle AEC
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There was no need for vertical angle arguments. Angle CDE is the exterior angle of triangle BED. Furthermore, angle AEC is the exterior angle of triangle BCE. So we can figure out their measures without looking at the vertical angles.
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This is fascinating. First, it makes me wonder: Is this true? That is, in an 80 degree - 80 degree isosceles triangle, does it really come out that swiping base-length segments up the triangle, will the last segment really leave a base-length segment on the side -- to really end up as this one has? Yes, it has to be true, because we can "build it" from the bottom and get the correct two blue angles of 20 and 140, meaning that the last angle will have to be 20, and thus it's isosceles. The smallest base angles would be 60, and that would allow for only a single triangle the same size as the original. If one started with 70 degree base angles, the first triangle would be 70-70-40, next would be 30-30-120, and the next would be....a miss. I suspect this would work only with 80-80 base angles...
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Hmm. If it started with 88 degree base angles then first triangle would be 88-88-4. Next would be 84-84-12, then 80-80-20, then 76, 72.... 4-4-172. It seems it would work. Perhaps it will work for even numbers above 80 degrees, where x,x,y and y is a factor of x (EDIT: which is only 88-88-4). How about 72-72-36, then 36-36-72 -- that would work for two triangles. Looks like 60, 72, 80, and 88 will work. Making 1 triangle, 2, 4, and 22? Too tired to check my work, lol.
I would suggest that
angle CDE can be directly calculated to be 3x, which is the exterior angle of the triangle EDB = 2x + x
similar case for angle AEC
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Awesome!! Sir, you are really genius. I love your tricks and skills. And for this I've been a fan of you. Take my greeting and lots of love.
Nice video ,thanks, dear professor
Thank you for this exciting video. I look forward to more. :-)
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Very well explained and fun to watch...!
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Very nice problem sir...thank you for uploading!!😃😃
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Very nicely explained. Thanks a lot.
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Thanks dear for this wonderful question.
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Interesting to use the exterior angles theorem. I guess it is just the combination of the interior angles theorem and a supplementary angle.
You are right on!
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Precisely. You just proved the theorem.
There was no need for vertical angle arguments. Angle CDE is the exterior angle of triangle BED. Furthermore, angle AEC is the exterior angle of triangle BCE. So we can figure out their measures without looking at the vertical angles.
Well done correct
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Excellent problem sir
Super video sir👍👍
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This is fascinating. First, it makes me wonder: Is this true? That is, in an 80 degree - 80 degree isosceles triangle, does it really come out that swiping base-length segments up the triangle, will the last segment really leave a base-length segment on the side -- to really end up as this one has? Yes, it has to be true, because we can "build it" from the bottom and get the correct two blue angles of 20 and 140, meaning that the last angle will have to be 20, and thus it's isosceles. The smallest base angles would be 60, and that would allow for only a single triangle the same size as the original. If one started with 70 degree base angles, the first triangle would be 70-70-40, next would be 30-30-120, and the next would be....a miss. I suspect this would work only with 80-80 base angles...
Awesome
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Superb
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Exterior Angle Theorem is confusing because hard to imagine the angles outside the triangle. Will you suggest a good web site of extended tutorial?
In triangle isocels ABC,we have AB=BCso,BCA=BAC so,CBA=angleX and CBA=180_(BCA+BAC)
*INTERESTING .....*
got it, just had to keep using base angles, opposite angles and exterior angles
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Thank for ptemath
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おおーー 素晴らしい問題だ
これはいい Great!😀
親愛なるあなたのとても素敵です！あなたは素晴らしいです👍あなたがそれを気に入ってくれてうれしいです！家族や友達とpremathチャンネルを共有し続けてください。親愛なる世話をし、祝福されたままでください😃
liked the video!
Thank you
A similar problem like this was asked in KVPY in India (Kvpy is exam)
今回は難しかったけど解けました。
お疲れ様でした。がんばり続ける👍
Интересный английский у парня😉
Hmm. If it started with 88 degree base angles then first triangle would be 88-88-4. Next would be 84-84-12, then 80-80-20, then 76, 72.... 4-4-172. It seems it would work. Perhaps it will work for even numbers above 80 degrees, where x,x,y and y is a factor of x (EDIT: which is only 88-88-4). How about 72-72-36, then 36-36-72 -- that would work for two triangles. Looks like 60, 72, 80, and 88 will work. Making 1 triangle, 2, 4, and 22? Too tired to check my work, lol.
Neat!
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Super O Super
There can be a simpler way of solving this question. Begin with the purple triangle..
Nice tip!
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So the yellow Triangle turns out to be an Equilateral Triangle.
You believe or not just 35 seconds and without pen & paper. Its easier than the easiest one. 20°
Very smart Emrah!
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@@PreMath you understand what i mean. 😉
yes
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x=20
Easy X = 20 degree