In figure PO $⊥$ QO. 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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Solution

$∠$ OPT = $∠$ OQT = $90^{\circ}$

PT = TQ (tangent property)

OP =OQ (radius of circle)
$∠$ PTQ = $90^{\circ}$ (angle sum property)

PTQO is a square(llgm with all angles 90 degree and adjacent sides are equal)
In a square the diagonals are ⊥ bisector
∴ PQ and OT are perpendiculaer bisector

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In figure PO ⊥ QO. 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.

∠ OPT =∠ OQT = 90∘ PT = TQ (tangent property) OP =OQ (radius of circle) ∠ PTQ = 90∘ (angle sum property) PTQO is a square(llgm with all angles 90 degree and adjacent sides are equal) In a square the diagonals are ⊥ bisector ∴ PQ and OT are perpendiculaer bisector

In figure PO $⊥$ QO. 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.