Question

In the alongside diagram, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B.

Prove that: 

1. AQ = BP
2. PQ = CD
3. ABPQ is a parallelogram.

Answer 1

Given: ABCD is a parallelogram and AP bisects ∠A and BQ bisects ∠B.

To prove:

1. AQ = BP
2. PQ = CD
3. ABPQ is a parallelogram.

Proofs:

Let ∠QAB = 2x

∠ABP = 180 - 2x

∠PAB = x

∠APB = 180 - (x + 180 - 2x)

= 180 - (180 - x)

= 180 - 180 + x

= x

In ΔBAP,

∠BAP = ∠APB

or AB = BP            ...(1)

Now, ∠PBA = 180 - 2x

∠PBQ = ∠QBA

∠QBA = `(180 - 2x)/2`

= 90 - x

So, ∠PBQ = 90 - x

and ∠PBQ = ∠AQB            ...[Alternate Interior Angles]

∠AQB = 90 - x

∠QBA = ∠AQB

hence, AB = AQ            ...(2)

From equations (1) & (2) we get

∴ AQ = PB

Since, AQ = BP &             ...(As ABCD is Parallelogram)

AQ || PB         ...(Opposite sides are equal & parallel, So, ABPQ is a parallelogram)

AB = PQ          ...(As ABPQ is a parallelogram)

and AB = CD           ...(As ABCD is a parallelogram)

Hence, CD = PQ