In the given figure , $P O ⊥ Q O$ . The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are the right bisectors of each other.

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Solution

From the given figure,

$O P = O Q$ (radius)
&
$∠ P O Q = 90^{0}$ (given) $\to (a)$

From tangent property, $T P = T Q \to (b)$

&

$∠ T P O = ∠ T Q O = 90^{0}$ $\to (c)$ {Tangent is always $⊥$ to Radii}

From $(a)$ , $(b)$ , $(c)$

$O P T Q$ form a square and bisector of square of square bisect each other at $90^{0}$ .

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In the given figure , PO⊥QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are the right bisectors of each other.

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In the given figure , $P O ⊥ Q O$ . The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are the right bisectors of each other.