Answer :

(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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