Question

Theorem-9.2 : The lengths of tangents drawn from an external point to a circle are equal.

Answer 1

Given : A circle with centre $O,P$ is a point outside the circle and $PA$ and $PB$ are two tangents to the circle from $P$. (See figure)

To prove : $PA=PB$

Proof : Join O, A, O, B and O, P.
$\angle OAP=\angle OBP={90}^{\circ }$

Now in the two right triangles $\left[\begin{array}{c}(\text{Angle between radii and tangents is}\\ {90}^{\circ }\text{according to theorem 9.1)}\end{array}\right]$

$△OAP$ and $△OBP$,
$OA=OB$ (radii of same circle)
$OP=OP$ (Common)

Therefore, by R.H.S. Congruency axiom,
$△OAP\cong △OBP$
This gives $PA=PB$
(CPCT)

Hence proved.