Question

solve this question​

Answer 1

Answer:

angle BPQ+angle DPQ=180

(sum of interior angle of same side to transversal)

1/2of angle BPQ=1/2of angle DQP=1/2×180

(BPR=RPQ and PQR=TQD)

RPQ+PQR=90 .... (1)

and

RPQ+PQR+QRP=180

( angle sum property of triangle)

90+QRP=180 (by eq. (1) )

QRP=180-90

QRP=90°

hence proved

Answer 2

If AB//CD

angle BPQ + DQP=180°

as given angle PQR is half of angle DQP

angle DQP =2 angle PQR

2(angle PQR + angle QPR)= 180°

angle PQR + angle QPR= 180/2

PQR + QPR= 90°

by angle sum property

angle PQR+ angle QPR+angle PRQ = 180°

as angle PQR + angle QPR = 90°

90 + angle PRQ=180

angle prq = 180-90

Angle PRQ is 90°

hence proved

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