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, 10:44 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 8th 2009, 12:35 AMukorov
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 AMthereddevils
- Nov 9th 2009, 06:59 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