Question

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

Answer 1

Answer:

We know that, the lengths of tangents drawn from an external point to a circle are equal.

∴ TP = TQ

In ΔTPQ,

TP = TQ

⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)

∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)

∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))

⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)

We know that, a tangent to a circle is perpendicular to the radius through the point of contact.

OP ⊥ PT,

∴ ∠OPT = 90º

⇒ ∠OPQ + ∠TPQ = 90º

⇒ ∠OPQ = 90º – ∠TPQ

⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)

From (1) and (2), we get

∠PTQ = 2∠OPQ

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Given- }}$

• A circle with centre O and TP and TQ are two tangents drawn to a circle from external point T.

$\large\underline{\sf{To\:prove - }}$

• ∠PTQ = 2 ∠OPQ

$\large\underline{\sf{Proof - }}$

• OP is radius of a circle

and

• PT is tangent to a circle.

As radius is perpendicular to tangent.

Therefore,

• ∠OPQ + ∠QPT = 90°

• ⇛ QPT = 90° - ∠OOQ -----(1)

Since, PT and QT are two tangents drawn to a circle from external point T.

We know,

Length of tangents drawn to a circle from external point are equal.

$\bf\implies \:TP = TQ$

Now,

$\rm :\longmapsto\: \sf \: In \: \triangle \: PTQ$

$\rm :\longmapsto\: \sf \: TP = TQ$

$\rm :\implies\: \angle \: QPT = \: \angle \: PQT$

$\: \: \: \: \{ \sf \: \because \: angle \: opposite \: to \: equal \: sides \: are \: equal \}$

$\rm :\implies\: \angle \: QPT = \: \angle \: PQT = 90 \degree \: - \angle \:OPQ$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{ \because \: \bf \: using \: (1) \}$

Again,

$\rm :\longmapsto\:In \: \triangle \: PTQ$

We know,

• Sum of angles of a triangle is 180°.

Therefore,

$\rm :\implies\: \angle \: QPT + \: \angle \: PQT + \angle \:PTQ = 180\degree \:$

$\rm :\longmapsto\:90\degree \: - \angle \:OPQ + 90\degree \: - \angle \:OPQ + \angle \:PTQ = 180\degree \:$

$\rm :\longmapsto\:180\degree \: - 2\angle \:OPQ + \angle \:PTQ = 180\degree \:$

$\rm :\longmapsto\: - 2\angle \:OPQ + \angle \:PTQ = 0$

$\bf\implies \:\angle \:PTQ = 2\angle \:OPQ$

${{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

• The tangent line never crosses the circle, it just touches the circle.

• At the point of tangency, it is perpendicular to the radius.

• A chord and tangent form an angle and this angle is same as that of tangent inscribed on the opposite side of the chord.

• From the same external point, the tangent segments to a circle are equal.