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 .

Printable View

- November 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 .

- November 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.

- November 9th 2009, 05:40 AMthereddevils
- November 9th 2009, 05:59 AMGrandad
Hello thereddevilsThere's nothing wrong with the question. Just use some simple trigonometry and Pythagoras' Theorem.

Let , and then if is the mid-point of :

(by symmetry)

Grandad