Q4 of 59 Page 58

In fig 3.27, the circles with centres P and Q touch each other at R. A line passing through R meets the circles at A and B respectively. Prove that -

(1) seg AP || seg BQ,


(2) ∆ APR ~ ∆ RQB, and


(3) Find ∠ RQB if ∠ PAR = 35°



(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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