1. ## isosceles triangle

In an isosceles triangle ABC , AB=AC and angle BAC =120 degree. Points D and E lie on BC such that BD=DE=EC . Prove that ADE is an equilateral triangle .

2. Using Sine Law, AD and AE can be proved to have equally divided angle BAC into 3 smaller ones, 40 degrees each. However, angles ADE and AED are 70 degrees (exterior angle of triangles), so triangle DAE should be isoscles, not equilateral.

3. Originally Posted by ukorov
Using Sine Law, AD and AE can be proved to have equally divided angle BAC into 3 smaller ones, 40 degrees each. However, angles ADE and AED are 70 degrees (exterior angle of triangles), so triangle DAE should be isoscles, not equilateral.
Thanks , so the problem is with the question . Are you really sure ??

4. Hello thereddevils
Originally Posted by thereddevils
In an isosceles triangle ABC , AB=AC and angle BAC =120 degree. Points D and E lie on BC such that BD=DE=EC . Prove that ADE is an equilateral triangle .
Originally Posted by ukorov
Using Sine Law, AD and AE can be proved to have equally divided angle BAC into 3 smaller ones, 40 degrees each. However, angles ADE and AED are 70 degrees (exterior angle of triangles), so triangle DAE should be isoscles, not equilateral.
Originally Posted by thereddevils
Thanks , so the problem is with the question . Are you really sure ??
There's nothing wrong with the question. Just use some simple trigonometry and Pythagoras' Theorem.

Let $\displaystyle AB= d$, and then if $\displaystyle M$ is the mid-point of $\displaystyle BC$:
$\displaystyle BD = \tfrac13BC = \tfrac23BM =\tfrac23d\cos30^o = \frac{d}{\sqrt3}$

$\displaystyle \Rightarrow DM = \tfrac12BD = \frac{d}{2\sqrt3}$
Also
$\displaystyle AM = d\sin30^o=\frac{d}{2}$
So using Pythagoras
$\displaystyle AD^2 = DM^2+AM^2 =\frac{d^2}{12}+\frac{d^2}{4}=\frac{d^2}{3}$

$\displaystyle \Rightarrow AD = \frac{d}{\sqrt3}= BD = DE = AE$ (by symmetry)
and hence $\displaystyle \triangle ADE$ is equilateral.