Given that PQRS is trapezium and PQ ∥SR AND PS∥QR
Let PQ is extended to T. Then, draw a line through R, which is parallel
to PS, intersecting PT at point T. It is clear that PTRS is a
(i) PS = RT (Opposite sides of parallelogram PTRS are equal)
However, PS = QR (Given)
Therefore, QR = RT
∠RTQ = ∠RQT (Angle opposite to equal sides are also equal)
Consider the parallel lines PS and RT. PT is the transversal line for them..
∠P + ∠RTQ = 180º ( Sum of angles on the same side are equal to 180 )
∠P + ∠RQT = 180º (Using the relation ∠RTQ = ∠RQT ) ... (1)
However, ∠Q + ∠RQT = 180º (Linear pair angles) ... (2)
From equations (1) and (2), we obtain,
∠P = ∠Q.