In the given figure, PO \[\perp\] QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OTare right bisector of each other.
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^o` (Given in the problem)
`∠OPT=90^o` (The radius of the circle will be perpendicular to the tangent at the point of contact)
`∠ TQO=90^o` (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^o`. Therefore,
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.