in a parallelogram ABCD the bisector of angle A meets DC in C and a b

1.	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
Answer» Answer: 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.

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

Answer»

Answer:

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.