Question

In the adjoining figure, the circles with centres $P$ and $Q$ touch each other at $RA$ line passing through $R$ meets the circles at $A$ and $B$ respectively. Prove that -
i. seg AP $\parallel$ seg $BQ$,
ii. $△APR\sim △RQB$, and
iii. Find $\angle RQB$ if $\angle PAR={35}^{\circ }$.

Answer 1

The circles with centres $P$ and $Q$ touch each other at $R$.
$\therefore$ By theorem of touching circles,
$P-R-Q$
i. In $△PAR$,
seg $PA=$ seg $PR$ [Radii of the same circle]
$\therefore \angle PRA\cong \angle PAR$ (i) [Isosceles triangle theorem]
Similarly, in $△QBR$,
seg $QR=\mathrm{seg}QB$ [Radii of the same circle]
$\therefore \angle RBQ\cong \angle QRB$ (ii) [Isosceles triangle theorem]
But, $\angle PRA\cong \angle QRB$ (iii) [Vertically opposite angles]
$\therefore \angle PAR\cong \angle RBQ$ (iv) [From (i) and (ii)]
But, they are a pair of alternate angles formed by transversal $AB$ on seg $AP$ and seg $BQ$.
$\therefore$ seg $AP\parallel$ seg $BQ$ [Alternate angles test]
ii. In $△APR$ and $△RQB$,
$\angle PAR\cong \angle QRB$ [From (i) and (iii)]
$\angle APR\cong \angle RQB$ [Alternate angles]
$\therefore △APR-△RQB[AA$ test of similarity]
iii. $\angle PAR={35}^{\circ }[$ Given] $\therefore \angle RBQ=\angle PAR={35}^{\circ }$ [From (iv)] In $△RQB$, $\angle RQB+\angle RBQ+\angle QRB={180}^{\circ }$ [Sum of the measures of angles of a triangle is ${180}^{\circ }$ ] $\therefore \angle RQB+\angle RBQ+\angle RBQ={180}^{\circ }$ [From (ii)] $\therefore \angle RQB+2\angle RBQ={180}^{\circ }$ $\therefore \angle RQB+2\times {35}^{\circ }={180}^{\circ }$ $\therefore \angle RQB+{70}^{\circ }={180}^{\circ }$ $\therefore \angle RQB={110}^{\circ }$