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.

Answer 1

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.