Question

In the following figure, P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. Prove that AD = 2CD.

Answer 1

Given in the question, in a parallelogram ABCD, P is a mid-point of BC such that ∠BAP = ∠DAP.

To prove that AD = 2CD

Proof: ABCD is a parallelogram

So, AD || BC and AB is transversal, then

∠A + ∠B = 180°   ...[Sum of cointerior angles is 180°]

∠B = 180° – ∠A   ...(i)

Now, in triangle ABP,

∠PAB + ∠B + ∠BPA = 180°   ...[By angle sum property of a triangle]

`1/2 ∠A + 180^circ - ∠A + ∠BPA = 180^circ`  ...[From equation (i)]

`∠BPA - (∠A)/2 = 0`

`∠BPA = (∠A)/2`   ...(ii)

∠BPA = ∠BAP

AB = BP  ...[Opposite sides of equal angles are equal]

In above equation multiplying both side by 2, we get

2AB = 2BP 

2AB = BC   ...[P is the mid-point of BC]

2CD = AD   ...[ABCD is a parallelogram, then AB = CD and BC = AD]