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.
Grandad