CBSE Class 9 Answered

SPQ = 60^o = QRS (opposite angles of a parallelogram)

PQR = PSR = 120^o (adjacent angles of a parallelogram)

As PA and QA bisect P and Q

1 = 30^o and 4 = 60^o

In PSA, 1 = 30^o , 2 = 120 ^o

So, using angle sum property, we get, 3 = 30^o

PS = SA ...(1)

(In a , sides opp. to equal angles are equal)

Similarly, QRA, 4 = 60^o and 6 = 60^o

QR = AR ...(2)

But PS = QR (opposite sides of a parallelogram) ...(3)

From (1), (2) and (3),

SA = AR

A is the mid-point of RS.