Question

In the given figure, 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°

Answer 1

(1) In ∆APR, AP = RP     (Radii of the same circle)
∴ ∠ARP = ∠RAP     .....(1)             (In a triangle, equal sides have equal angles opposite to them)     
In ∆BQR, QR = QB     (Radii of the same circle)
∴ ∠RBQ = ∠BRQ    .....(2)             (In a triangle, equal sides have equal angles opposite to them)
Also, ∠ARP = ∠BRQ          .....(3)          (Vertically opposite angles)
From (1), (2) and (3), we have
∠RAP = ∠RBQ
∴ seg AP || seg BQ    (If a transversal intersect two lines such that a pair of alternate interior angles is equal, then the two lines are parallel)
(2) In ∆APR and ∆RQB,
∠RAP = ∠RBQ       (Proved)
∠ARP = ∠BRQ        (Vertically opposite angles)

∴ ∆APR ~ ∆RQB     (AA similarity criterion)
(3) ∠RBQ = ∠PAR = 35º
∴ ∠BRQ = ∠RBQ = 35º
In ∆RQB,
∠BRQ + ∠RBQ + ∠RQB = 180º     (Angle sum property)
∴ 35º + 35º + ∠RQB = 180º
⇒ 70º + ∠RQB = 180º
⇒ ∠RQB = 180º − 70º = 110º
Thus, the measure of ∠RQB is 110º.