In the adjoining figure, the circles with centres P and Q touch each o

1.	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» 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°

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»

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°