Question

In the given figure, $XY$ and $X^{\prime }{Y}^{\prime }$ are two parallel tangents to a circle with center $O$ 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

$\angle AOB={90}^o$

Join OC $OC\perp AB$

(Tangent at any pt of circle is far to radius)

$\angle ACO=\angle BCO={90}^o$

In $\Delta AOP$ & $\Delta AOC$

$OP=OC$

$AP=AC$ (length of drawn from external point to circle are equal)

$\Rightarrow \Delta AOP\cong \Delta AOC$

In $\Delta BOC$ & $\Delta BOQ$

with same argument

$\Delta BOC\cong \Delta BOQ$

$\angle BOC=\angle BOQ$ (CPCT)

for line PQ

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

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

$\Rightarrow \angle AOC+\angle BOC=90$

$\Rightarrow \angle AOB=90$