Question

In the given figure, $XY$ and $X^{\prime }{Y}^{\prime }$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X^{\prime }{Y}^{\prime }$ at $B$. Prove that $\angle AOB={90}^{\circ }$.

Answer 1

In $△AOP$ & $△AOC$

$OP=OC$ [Both radius]

$AP=AC$ [Length of tangents drawn from external point to a circle are equal]

$OA=OA$ [common]

$\therefore △AOC\cong △AOP$ [SSS Congruence rule]

So, $\angle AOP=\angle AOC...............\left(1\right)$ [CPCT]

Now,

In $△BOC$ & $△BOQ$

$OC=OQ$ [Both radius]

$BC=BQ$ [Length of tangents drawn from external point to a circle are equal]

$OB=OB$ [common]

$\therefore △BOC\cong △BOQ$ [SSS Congruence rule]

So, $\angle BOC=\angle BOQ...............\left(2\right)$ [CPCT]

For line PQ

$\angle AOP+\angle AOC+\angle BOC+\angle BOQ=180$

$\angle AOC+\angle AOC+\angle BOC+\angle BOC=180$

$2\angle AOC+2\angle BOC=180$

$2(\angle AOC+\angle BOC)=180$

$\angle AOC+\angle BOC=\frac{180}{2}$

$\angle AOC+\angle BOC=90$

$\angle AOB=90$

Hence proved.