Q. 244.4( 9 Votes )

# In Fig. 10.62, . The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

Answer :

**To prove:**PQ and OT are the right bisectors.

**Proof:**

To prove PQ and OT are the right bisectors,

We need to prove ∠PRT= ∠TRQ=∠QRO=∠ORP = 90

^{º}

As it is given that ,

⇒ ∠POQ = 90

^{º}

In Δ POT and Δ OQT

OP=OQ (Radius)

∠OPT = ∠OQT = 90

^{º}( Tangent to a circle at a point is perpendicular to the radius through the point of contact)

OT=OT (common)

∴ Δ POT ≅ Δ OQT

Thus PT=OQ ( BY C.P.C.T) ..... (1)

Now in Δ PRT and Δ ORQ

∠TPR = ∠OQR ( alternate angles)

∠PTO = ∠TOQ (alternate angles)

PT=OQ ( from (1) )

∴ Δ PRT ≅ Δ ORQ

Thus TQ = OP ( By C.P.C.T 0

Hence PT=TQ=OQ=OP

Thus it is a square,

⇒ The diagnols bisect at 90

^{º}.

Hence proved

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PREVIOUSIn Fig. 10.61, BDC is a tangent to the given circle at point D such that BD = 30 cm and CD = 7 cm. The other tangents BE and CF are drawn respectively from B and C to the circle and meet when produced at A making BAC a right angle triangle. Calculate (i) AF (ii) radius of the circle.NEXTIn Fig. 10.63, two tangents AB and AC are drawn to a circle with centre O such that . Prove that OA = 2AB.

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