If a triangle has two angle bisectors of equal length, is it an isoceles triangle? I heard this problem on a 3B1B interview with Steven Strogatz (https://www.youtube.com/watch?v=SUMLKweFAYk&t=1048s , 2:44)
I was hoping someone could let me know whether my sketch proof is valid, if I have misunderstood, or suggest their version of a proof?
Sketch Proof:
On the cartesian plane, draw a line connecting the points (-1/2, 0) and (1/2,0). Draw two lines from each points to a third at (x, y). All euclidean triangles are similar to the triangle of this construction with some value (x,y). If y is fixed: as x tends to -inf the bisector of (-1/2,0) goes to length 0 and of (1/2,0) goes to +inf. As x tends to +inf this is reversed. The bisector lengths are strictly monotonic with respect to x. Their difference is strictly monotonic with respect to x, therefore their difference is only zero when x = 0. When their difference is zero the bisectors are the same length, so are congruent. All triangles similar to the construction for some y and x = 0 are isosceles. Therefore all triangle who have two congruent bisectors are isosceles.
 A: Good video. I watched it too and went right away to solve the problem. I do not know if your version of the proof is correct, but this is my proof using only euclidian geometry. Hope it helps to give you new ideas (I think maybe the proof young Steven did could be similar). Sorry, I had to attach it all in an image, I'm new to the platform and don´t know how to write the symbols.

A: Reference: Encyclopedia of elementary geometry.
History: In 1840 Lehmus asked Steiner to solve this problem. Steiner gave a complicated solution.This caused many mathematicians look for easier solutions. From 1842 many solutions were proposed for this problem. One was the solution gave by French engineer Descub which is posted by user MMMagician. Another solution is given in reference I derived this history from.Here is the second solution:

Draw the bisector of angle $\hat A$ which passes O ,the intersection of two other bisectors $BD=CE$.As shown in figure  we construct triangle A'B'C' equal to triangle ABC such that it's bisector C'E' of angle $\hat C'$ is coincident on the bisector BD of angle $\hat B $ of triangle ABC. Connect A to A', quadrilateral BDAA' is cyclic because $\widehat{BA'D}=\widehat{BAD}$. So $\widehat{ABD}=\widehat{AA'D}$. In triangle AOB we have:
$\widehat{AOD}=\widehat{OBA}+\widehat{OAB}=\alpha+\beta=\widehat{O'A'D}+\widehat{DA'A}=\widehat{O'A'A}$
where $\alpha=\frac{\widehat{BAC}}2$ and $\beta=\frac{\widehat{ABC}}2$
Therefore quadrilateral OO'A'A is also cyclic, since $OA=OA'$ it is also isosceles. In this way $OO'||AA'$ and $\widehat{OAA'}=\widehat{O'A'A}$ , so AA'BD is also isosceles trapezoid, that is it's diagonals are equal i.e $A'D=AB$. But $A'D=A'C'=AC$ which results in $AB=AC$ that is triangle ABC is isosceles.
