Question

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

Answer 1

Answer:

Refer image

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

o

or, ∠OPQ+∠TPQ=90

o

or, ∠TPQ=90

o

−∠OPQ−−−(3)

In △PTQ

∠TPQ+∠PQT+∠QTP=180

o

(∴ Sum of angles triangle is 180

o

)

or, 90

o

−∠OPQ+∠TPQ+∠QTP=180

o

or, 2(90

o

−∠OPQ)+∠QTP=180

o

[from (2) and (3)]

or, 180

o

−2∠OPQ+∠PTQ=180

o

∴ 2∠OPQ=∠PTQ−−−− proved

Answer 2

Given,

a circle with centre O, an external point T and two tangents TP and TQ to the cirlce

$\:$

where,

P, Q are the points of contact (See figure)

$\:$

We need to prove that,

∠PTQ = 2∠OPQ

$\:$

Let,

∠PTQ = θ

$\:$

Now,

by Theorem, TP = TQ

$\:$

So,

ΔTPQ is an isosceles triange

$\:$

Therefore,

∠TPQ + ∠PQT + ∠QTP = 180° (Sum of three angles)

$\displaystyle \sf⟹ \: ∠TPQ + ∠PQT \: = \: \frac{1}{2} (180 \degree - \theta) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: 90 \degree - \frac{1}{2} \theta$

$\:$

Also,

by Theorem, ∠QPT = 90°

$\:$

So,

$\sf∠OPQ = ∠OPT - ∠TPQ$

$\displaystyle\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 90 \degree - \bigg[90 - \frac{1}{2} \theta \bigg ] \: = \: \frac{1}{2} \theta \: = \: \frac{1}{2} \angle \: PTQ$

$\:$

This gives,

$\displaystyle \sf∠OPQ \: = \: \frac{1}{2} ∠PTQ$

$∴ \: \displaystyle \sf ∠PTQ \: = \: 2∠OPQ$

Similarly ∠PTQ = 2∠OQP

$\:$

Hence proved !!

$\:$