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Question
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 right bisector of each other.
Solution
In the given figure,
PO = OQ (Since they are the radii of the same circle)
PT = TQ (Length of the tangents from an external point to the circle will be equal)
Now considering the angles of the quadrilateral PTQO, we have,
∠POQ=90° (Given in the problem)
∠OPT=90° (The radius of the circle will be perpendicular to the tangent at the point of contact)
∠TQO=90° (The radius of the circle will be perpendicular to the tangent at the point of contact)
We know that the sum of all angles of a quadrilateral will be equal to 360°. Therefore,
∠POQ+∠TQO+∠OPT+∠PTQ=360°
90°+90°+90°+∠PTQ=360°
∠PTQ=90°
Thus we have found that all angles of the quadrilateral are equal to 90°.
Since all angles of the quadrilateral PTQO are equal to 90° and the adjacent sides are equal, this quadrilateral is a square.
We know that in a square, the diagonals will bisect each other at right angles.
Therefore, PQ and OT bisect each other at right angles.
Thus we have proved.
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