Let ABCD be a quadrilateral, and let AR, BP, CP, and DR be the angle bisectors of ∠A, ∠B, ∠C, and ∠D respectively. These bisectors intersect to form a quadrilateral PQRS. To Prove: PQRS is a cyclic quadrilateral. **Proof:** In triangle △ARD: \[ \angle ARD + \angle RDA + \angle DAR = 180^\circ \Rightarrow \angle ARD + \frac{1}{2}\angle A + \frac{1}{2}\angle D = 180^\circ \quad \text{(i)} \] In triangle △BPC: \[ \angle BPC + \angle PCB + \angle CBP = 180^\circ \Rightarrow \angle BPC + \frac{1}{2}\angle C + \frac{1}{2}\angle B = 180^\circ \quad \text{(ii)} \] Adding equations (i) and (ii): \[ \angle ARD + \angle BPC + \frac{1}{2}(\angle A + \angle B + \angle C + \angle D) = 360^\circ \Rightarrow \angle ARD + \angle BPC + \frac{1}{2} \times 360^\circ = 360^\circ \Rightarrow \angle ARD + \angle BPC = 180^\circ \] So, ∠QRS + ∠QPS = 180°, which means a pair of opposite angles in quadrilateral PQRS are supplementary.