Let PQRS be a quadrilateral in which ∠PQR and ∠PSR are supplementary, i.e., \[ ∠PQR + ∠PSR = 180^\circ \] We need to prove that the vertices P, Q, R, and S lie on a circle (i.e., the quadrilateral is cyclic). **Construction:** A circle can be drawn through any three non-collinear points P, Q, and R. Suppose the circle does not pass through point S. Let the circle intersect PS or its extension at point T. Join T and R. **Proof:** According to the construction, PQRT is a cyclic quadrilateral. So, \[ ∠PQR + ∠PTR = 180^\circ \quad \text{(1)} \] But it is given that \[ ∠PQR + ∠PSR = 180^\circ \quad \text{(2)} \] From (1) and (2), we get: \[ ∠PTR = ∠PSR \] But in a triangle, an exterior angle cannot be equal to the opposite interior angle. So, ∠PTR = ∠PSR only when points S and T coincide. Therefore, the circle passing through P, Q, and R must also pass through S. Hence, the points P, Q, R, and S lie on the same circle.