Question

In the adjoining 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 –

i. seg AP || seg BQ, 

ii. ∆APR ~ ∆RQB, and 

iii. Find ∠RQB if ∠PAR = 35°.

Answer 1

The circles with centres P and Q touch each other at R. 

∴ By theorem of touching circles, 

P – R – Q 

i. In ∆PAR, seg PA = seg PR [Radii of the same circle]

∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem] 

Similarly, in ∆QBR, 

seg QR = seg QB [Radii of the same circle] 

∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem] 

But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles] 

∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)] 

But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ.

∴ seg AP || seg BQ [Alternate angles test] 

ii. In ∆APR and ∆RQB, 

∠PAR ≅ ∠QRB [From (i) and (iii)] ∠APR ≅ ∠RQB [Alternate angles]

∴ ∆APR – ∆RQB [AA test of similarity] 

iii. ∠PAR = 35° [Given] 

∴ ∠RBQ = ∠PAR= 35° [From (iv)] 

In ∆RQB, 

∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°] 

∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)] 

∴ ∠RQB + 2 ∠RBQ = 180° 

∴ ∠RQB + 2 × 35° = 180° 

∴ ∠RQB + 70° = 180° 

∴ ∠RQB = 110°