Question

Question 9
In Fig. 10.13, XY and X′Y′ 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′Y′ at B. Prove that $\angle AOB={90}^{\circ }$.

Answer 1

We joined O and C.
According to question,
In $\Delta$ OPA and $\Delta$ OCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
$\therefore \Delta OPA\cong \Delta OCA$ (SSS congruence criterion)
$\Rightarrow \angle POA=\angle COA$ … (i)
Similarly,
$\Delta OQB\cong \Delta OCB$
$\angle QOB=\angle COB$ … (ii)
Since POQ is a diameter of the circle, it is a straight line.
$\therefore \angle POA+\angle COA+\angle COB+\angle QOB={180}^{\circ }$
From equations (i) and (ii),
$2\angle COA+2\angle COB={180}^{\circ }$
$\Rightarrow \angle COA+\angle COB={90}^{\circ }$
$\Rightarrow \angle AOB={90}^{\circ }$