Thread: 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 .

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 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 ??

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 , and then if is the mid-point of :

Also

So using Pythagoras

(by symmetry)

and hence is equilateral.

Grandad