in a parallelogram ABCD the bisector of angle A meets DC in C and a b is equal to 280 prove that BD bisects Angle B
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Let AD=x,AB=2,AD=2x
Also AP is the bisector ∠A∴∠1=∠2
Now, ∠2=∠5 (alternate angles)
∴∠1=∠5NowAD=DP=x [∵ Sides opposite to equal angles are also equal]
∵AB=CD (opposite sides of parallelogram are equal)
∴CD=2x⇒DP+PC=2x⇒x+PC=2x⇒PC=x
Also,BC=x in ΔBPC,∠6=∠4 (Angles opposite to equal sides are equal)
Also, ∠6=∠3 (alternate angles)
∵∠6=∠4 and ∠6=∠3⇒∠3=∠4
Hence, BP bisects ∠B.
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