Question

From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.

Answer 1

$\huge\sf\blue{Given}$

✭ T is a point outside a circle of center O

✭ TP & TQ are tangents

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$\huge\sf\gray{To \: Prove}$

◈ OT is the right bisector of line segment PQ

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$\huge\sf\purple{Steps}$

We see that in ∆PTR & ∆QTR

➝ $\sf TP=TQ \:\:\:\:$ «« Tangents of the circle from the same point »»

➝ $\sf TR = TR \:\:\:\:$ «« Common »»

➝ $\sf \angle TPR = \angle TAR$ «« Opposite angles of equal sides are equal »»

➝ $\sf \triangle PTR \cong \triangle QTR$ «« SAS Congruency »»

So therefore

➳ $\sf PR = QR$ «« CPCT »»

➳ $\sf \angle PRT = \angle QRT$

We see that PQ is a straight line,

➳ $\sf \angle PRT + \angle QRT = 180^{\circ}$

➳ $\sf \angle PRT + \angle PRT = 180^{\circ}$

➳ $\sf 2\angle PRT = 180^{\circ}$

➳ $\sf PRT = \dfrac{180}{2}$

➳ $\sf \red{PRT = 90^{\circ}}$

➳ $\sf \orange{QRT = 90^{\circ}}$

$\sf Hence \ Proved!!$

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Answer 2

Given :-

• A circle with centre O.
• Tangents TP and TQ are drawn from a point T outside a circle.

To Prove :-

• OT is the right bisector of line segment PQ.

Construction :-

• Join OP & OQ

Proof :-

• In ΔPTR and ΔQTR
• In ΔOPT and ΔOQT

∠OPT = ∠OQT = 90°

OP = OQ (radius)

OT = OT (Common)

ΔOPT ≅ ΔOQT (By RHS congruence)

∠PTR = ∠QTR (cpct)

TP = TQ (Tangents are equal)

TR = TR (Common)

∠PTR = ∠QTR (OT bisects ∠PTQ)

ΔPTR ≅ ΔQTR (By SAS congruency)

PR = QR

∠PRT = ∠QRT

But ∠PRT+ ∠QRT = 180° (as PQ is line segment)

∠PRT = ∠QRT = 90°

Therefore, TR or OT is the right bisector of line segment PQ$.$