Question

In a triangle ABC, AB = AC. A circle passing through B and C intersects the side AB and AC at D and E respectively. Prove that DE is parralel to BC.

Answer 1

Given,

AB=AC

Hence angle ACB = angle ABC =x

Now in quad DECB

Angle BDE + Angle ECB = 180 (cyclic quadrilateral)

AngleBDE + x =180

AngleBDE=180-x

Similarly

AngleDEC=180-x

Hence as Angle DBC +BDE = x +180–x=180

DE//BC(adjacent angles are supplementary)