Question

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

Answer 1

To Prove:-

• ∠PTQ=2 ∠OPQ.

Proof:-

• Refer image

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

$\:\:\:$∴ TP=TQ−−−(1)

∴ ∠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°

• → ∠OPQ + ∠TPQ=90°

• ∠TPQ = 90° − ∠OPQ −−−(3)

$\:\:\:$ In △PTQ

• → ∠TPQ + ∠PQT + ∠QTP = 180° (∴ Sum of angles triangle is 180° )

• → 90° − ∠OPQ + ∠TPQ + ∠QTP = 180

• → 2(90° − ∠OPQ ) +∠QTP = 180°

$\:\:\:$ [from (2) and (3)]

• → 180° − 2∠OPQ + ∠PTQ = 180 °

$\:\:\:$ ∴ 2∠OPQ=∠PTQ

$\:\:\:$ proved!!