Given: A cyclic quadrilateral ABCD in which the angle bisectors AR, BR, CP and DP of internal angles A, B, C and D respectively form a quadrilateral PQRS.
To prove: PQRS is a cyclic quadrilateral.
Proof:
In △ARB, we have:
(i) 1/2∠A + 1/2∠B + ∠R = 180° (Since, AR, BR are bisectors of ∠A and ∠B respectively)
In △DPC, we have:
(ii) 1/2∠D + 1/2∠C + ∠P = 180° (Since, DP,CP are bisectors of ∠D and ∠C respectively)
Adding (i) and (ii),we get:
1/2∠A + 1/2∠B + ∠R + 1/2∠D + 1/2∠C + ∠P = 180° + 180°
∠P + ∠R = 360° - 1/2 x (∠A + ∠B + ∠C + ∠D)
∠P + ∠R = 360° - 1/2 x 360° = 360° - 180°
⇒ ∠P + ∠R = 180°
As the sum of a pair of opposite angles of quadrilateral PQRS is 180°, quadrilateral PQRS is cyclic.
