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) ΔAPR ~ ΔRQB, and (3) Find ∠RQB if ∠PAR = 35°

Answer: (1) and (2) are proved below; (3) \(\angle RQB = 110^\circ\).

By the theorem of touching circles, P, R, Q are collinear.

(1) \(\angle PRA \cong \angle QRB\) (vertically opposite angles).

PA = PR (radii), so \(\angle PAR \cong \angle PRA\). QR = QB (radii), so \(\angle QRB \cong \angle QBR\).

Hence \(\angle PAR \cong \angle QBR\). These are alternate angles made by transversal AB with segments AP and BQ, so seg AP ∥ seg BQ.

(2) In \(\triangle APR\) and \(\triangle RQB\):

• \(\angle PAR \cong \angle QRB\) (both equal \(\angle PRA\))
• \(\angle PRA \cong \angle QBR\) (both equal \(\angle QRB\))

\(\therefore \triangle APR \sim \triangle RQB\) (AA test), with A ↔ R, P ↔ Q, R ↔ B.

(3) \(\angle PAR = \angle PRA = 35^\circ\), so

\(\angle RQB\) corresponds to \(\angle APR\), so \(\angle RQB = 110^\circ\).