Question

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.

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,

${\displaystyle \angle POQ={90}^o}$ (Given in the problem)

${\displaystyle \angle OPT={90}^o}$ (The radius of the circle will be perpendicular to the tangent at the point of contact)

${\displaystyle \angle TQO={90}^o}$ (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 ${\displaystyle {360}^o}$. Therefore,

${\displaystyle \angle POQ+\angle TQO+\angle OPT+\angle PTQ={360}^o}$

${\displaystyle {90}^O+{90}^O+{90}^O+\angle PTQ={360}^o}$

${\displaystyle \angle PTQ={90}^o}$

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.