Question

In the figure XY and X'Y' are two parallel tangents to a circle with centre O and and another tangent AB with point of contact C interesting XY at A and X'Y' at B prove that $\angle AOB={90}^0$.

Answer 1

Consider the problem

Let us join point $O$ to $C$

In $\Delta OPAand\Delta OCA$

$OP=OC$ (Radii of the same circle)

$AP=AC$ (Tangent from point $A$)

$AO=AO$ (Common side)

$\Delta OPA\cong \Delta OCA$ ($SSS$ congruence criterion)

Therefore, $P\leftrightarrow C,A\leftrightarrow A,O\leftrightarrow O$

$\angle POA=\angle COA.........\left(1\right)$

Similarly,

$\angle QOB\cong \angle OCB$

$\angle QOB=\angle COB.........\left(2\right)$

Since,$POQ$ is a diameter of the circle, it is a straight line.

Therefore, $\angle POA+\angle COA+\angle COB+\angle QOB={180}^{\circ }$

So, from equation $\left(1\right)$ and equation $\left(2\right)$

$\begin{array}{c}2\angle COA+2\angle COB={180}^{\circ }\\ \angle COA+\angle COB={90}^{\circ }\\ \angle AOB={90}^{\circ }\end{array}$