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In Fig. \(PO\perp OQ.\) 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.

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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 = 90º

As it is given that \(PO\perp OQ,\)

⇒ ∠POQ = 90º 

In Δ POT and Δ OQT

OP = OQ (Radius)

∠OPT = ∠OQT = 90º 

(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

Hence PT=TQ=OQ= OP 

Thus it is a square,

⇒ The diagnols bisect at 90º .

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