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 .

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- Nov 7th 2009, 09:44 PMthereddevilsisosceles 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 .

- Nov 7th 2009, 11:35 PMukorov
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.

- Nov 9th 2009, 05:40 AMthereddevils
- Nov 9th 2009, 05:59 AMGrandad
Hello thereddevilsThere'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 $\displaystyle BC$:$\displaystyle BD = \tfrac13BC = \tfrac23BM =\tfrac23d\cos30^o = \frac{d}{\sqrt3}$Also

$\displaystyle \Rightarrow DM = \tfrac12BD = \frac{d}{2\sqrt3}$

$\displaystyle AM = d\sin30^o=\frac{d}{2}$So using Pythagoras$\displaystyle AD^2 = DM^2+AM^2 =\frac{d^2}{12}+\frac{d^2}{4}=\frac{d^2}{3}$and hence $\displaystyle \triangle ADE$ is equilateral.

$\displaystyle \Rightarrow AD = \frac{d}{\sqrt3}= BD = DE = AE$ (by symmetry)

Grandad