Question

Example 2: Two tangents TP and TQ are drawn
to a circle with centre O from an external point T.
Prove that Z PTQ = 2 Z OPQ.​

Answer 1

We know that length of taughts drawn from an external point to a circle are equal.....

$∴ TP=TQ−−−(1)$

$4 \: ∴ ∠TQP \: = \: ∠TPQ \: \\ \\ (angles \: of \: equal sides \: are \: equal) \\ −−−(2)$

Now, PT is tangent and OP is radius.........

$∴  OP \: ⊥ \: TP \\ \\ (Tangent \: at \: any \: point \: pf\\ \:circle \: is \: perpendicular \: to \: the\\ \: radius \: through \: point \: of \: cant \: act)$

$∴ ∠OPT=90$

$or \\ ∠OPQ+∠TPQ= {90}^{0}$

$or\\ ∠TPQ=90^{0} −∠OPQ−−−(3)$

$In △PTQ$

$∠TPQ \: + \: ∠PQT \: + \: ∠QTP \\ = 180^{0} \\ (Sum \: of \: angles \: triangle \: is \:  180^{0})$

$or,  \\ 90^{0} −∠OPQ+∠TPQ+∠QTP \\ =180$

$or, \\  2(90^{0} −∠OPQ)+∠QTP=180^{0} \\  [from (2) and (3)]$

$or,  \\ 180^{0} −2∠OPQ+∠PTQ=180^{0}$

$∴ 2∠OPQ=∠PTQ \: \\ −−−−  proved$

HOPE IT HELPS YOU....