Question

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

Answer 1

Answer:

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.