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We assume the quadrilateral PQRS has sum of opposite angles \[{180^ \circ }\]

\[\angle PSR + \angle PQR = {180^ \circ }\]

\[\angle SRQ + \angle SPQ = {180^ \circ }\]

We know there are three non-collinear points P, S, and R. We know a circle with centre O passes through three collinear points, so we assume point Q does not lie on Circle, so three collinear points are P, Q” and Q.

PQ”RS is a cyclic quadrilateral as all the vertices of the quadrilateral lie on the circle. We know that for a cyclic quadrilateral sum of opposite angles is ${180^ \circ }$. That is

\[\angle PSR + \angle PQ''R = {180^ \circ }\] … (1)

But it is given

\[\angle PSR + \angle PQR = {180^ \circ }\] … (2)

So by comparing (1) and (2) we can write

\[\angle PSR + \angle PQR = \angle PSR + \angle PQ''R\]

\[\angle PSR - \angle PSR + \angle PQR = \angle PQ''R\]

\[\angle PQR = \angle PQ''R\] … (3)

In triangle RQQ”, from exterior angle property which states that the exterior of the triangle is equal to the sum of opposite angles of the triangle.

\[\angle PQ''R = \angle Q''QR + \angle Q''RQ\] … (4)

Using Equation (3) in Equation (4)

\[\angle PQR = \angle Q''QR + \angle Q''RQ\]

We can see from the diagram that

\[

\angle PQR = \angle PQR + \angle Q''RQ \\

\angle PQR - \angle PQR = \angle Q''RQ \\

0 = \angle Q''RQ \\

\]

This can be only true if, Q and Q” coincides. Therefore Point Q should lie on the circle.

Hence it is proved that if the sum of the opposite angles of a quadrilateral is \[{180^ \circ }\] , the quadrilateral is cyclic.