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
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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