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.