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 and X'Y' at B. Prove that $\angle AOB={90}^o$.

Answer 1

In $△OPA$ and $△OCA$

$m\angle P\cong m\angle C$ ...(Tangent is $\perp$ to radius)

$segPA\cong segAC$ ...(Tangent from external point)

$segOA\cong segOA$ ...(common side)

$\therefore △OPA\cong △OCA$ ...(RHS test of congruence)

$\angle POA=\angle COA=x$ ...(c.a.c.t) .....(1)

Similarly, $△OCB\cong △OQB$.

$\angle COB=\angle QOB=y$ ...(c.a.c.t) ....(2)

Since $P-O-Q$, $\angle POQ={180}^o$

$\angle POA+\angle COA+\angle COB+\angle QOB={180}^o$

(Angle addition property)

$\therefore x+x+y+y={180}^o$

$\therefore 2x+2y={180}^o$

$\therefore x+y={90}^o$

$\angle AOB={90}^o$