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ABC is a isosceles triangle triangle in which AB= AC . A circle,passes through B and C intersects the sides AB and AC at D and E respectively. Prove that

DE is parallel to BC



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Isosceles triangle  =>  angle B = angle C

The points BCED are on the circle.  Hence BCED is a cyclic quadrilateral.

Hence opposite angles are supplementary. 

angle C + angle BDE = 180 deg

=>  angle B + angle BDE = 180 deg

=>  angle B = 180 - angle BDE = Angle ADE  

Hence  DE is parallel to  BC

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