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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 *OT*are right bisector of each other.

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#### 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^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,

`∠ POQ+∠TQO+∠OPT+∠PTQ=360^o`

`90^O+90^O+90^O+∠ PTQ=360^o`

`∠ 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.

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