Question

P is the midpoint of side BC of parallelogram ABCD such that angle BAP= angle DAP. Prove that AD= 2 times CD.

Answer 1

Given : ABCD is a parallelogram. P is the mid point of BC and ∠BAP = ∠DAP

To prove : AD = 2 CD

Proof : Given, ∠BAP = ∠DAP
∴ ∠1 = ∠BAP = 1/2 ∠A  ...(1)

ABCD is a parallelogram,
∴ AD || BC  (Opposite sides of the parallelogram are equal)
∠A + ∠B = 180°  (Sum of adjacent interior angles is 180°)
∴ ∠B = 180° – ∠A  ...(2)

In ΔABP,
∠1 + ∠2 + ∠B = 180° (Angle sum property)
=> 1/2∠A + ∠2 + 180 - ∠A = 180 [Using equations (1) and (2)]
=> ∠A - 1/2 ∠A = 0
=> ∠A = 1/2 ∠A ...(3)
From (1) and (2), we have
∠1 = ∠2
In ΔABP,
∠1 = ∠2
∴ BP = AB   (In a triangle, equal angles have equal sides opposite to them)
=> 1/2 BC = AB (P is the midpoint on BC)
=> BC = 2AB
⇒ AD = 2CD  (Opposite sides of the parallelogram are equal)

Hence, proved.

Answer 2

Pic have the solution