Question

Prove that the bisectors of the interior angles of a rectangle form a square.

Answer 1

Given: A parallelogram ABCD in which AR, BR, CP, DP are the bisects of ∠A, ∠B, ∠C, ∠D, respectively forming quadrilaterals PQRS.

To prove: PQRS is a square.

Proof:

In Δ ARB,

∠RAB + ∠RBA + ∠ARB = 180°

45° + 45° + ∠ARB = 180°

90° + ∠ARB = 180°

∠ARB = 180° - 90°

∴ ∠ARB = 90°

Similarly, ∠SRQ = 90°

In Δ ARB,

AR = BR  ...(i)

ΔASD ≅ Δ BQC   ...[By ASA rule]

AS = BQ  ...(ii)  [by CPCTC]

(i) - (ii)

AR - AS = BR - BQ

SR = RQ   ...(iii)

Also, SP = PQ  ...(iv)

PQ = RS  ...(v)

Hence, PQRS is a square.