# isosceles triangle

• Nov 7th 2009, 10:44 PM
thereddevils
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 .
• Nov 8th 2009, 12:35 AM
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.
• Nov 9th 2009, 06:40 AM
thereddevils
Quote:

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 ??
• Nov 9th 2009, 06:59 AM
Hello thereddevils
Quote:

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 .

Quote:

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.

Quote:

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

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

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