Question

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

Answer 1

Given, XY $\&$ X'Y' are parallel

AB is another tangent which touches the circle at C.



To prove: $\angle$ AOB = ${90}^{\circ }$

Const. : Join OC.

Proof : In $△$ OPA and $△$ OCA

OP = OC $\left(Radii\right)$

$\angle$ OPA = $\angle$ OCA $(Radius\perp tangent)$

OA = OA $\left(Common\right)$

$\therefore △OPA\cong △OCA\left(RHSCongRule\right)$

$\angle 1=\angle 2.........\left(i\right)$

Similarly, $△OQB\cong △OCB$

$\angle 3=\angle 4....\left(ii\right)$

But, POQ is a diameter of circle

$\angle POQ={180}^{\circ }\left(Straightangle\right)$

$\angle 1+\angle 2+\angle 3+\angle 4={180}^{\circ }$

From eq. (i) and (ii)

$\angle 2+\angle 2+\angle 3+\angle 3={180}^{\circ }$

$2(\angle 2+\angle 3){180}^{\circ }$

$\angle 2+\angle 3={180}^{\circ }$

$\angle 2+\angle 3={90}^{\circ }$

Hence, $\angle AOB={90}^{\circ }$
Hence proved.