# In fig 3.27, the

(1)In ΔAPR,

AP = RP {Radius of the circle with centre P}

PAR = PRA … (1)

In ΔRQB,

RQ = QB {Radius of the circle with centre Q}

QRB = QBR …. (2)

PRA = QRB {Vertically Opposite Angle} ….(3)

PAR = QBR {From (1), (2) and (3)}

Alternate interior angles are equal.

AP || BQ

Hence, proved.

(2) In ∆ APR and ∆ RQB,

PAR = QBR and PRA = QRB {From (1) and (2)}

∆ APR ~ ∆ RQB {AA}

Hence, proved.

(4) Given: PAR = 35°

QBR = 35° = QRB {Proved previously}

In ∆ RQB,

RQB + QRB + QBR = 180° {Angle sum property of the triangle}

⇒∠ RQB + 35° + 35° = 180°

⇒∠ RQB = 180°- 70° = 110°

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