# Two radii OA and OB of a circle with centre O are perpendicular to each other. If two tangents drawn at the point A and B intersect each other at the point T, let us prove that AB = OT and they bisect each other at a right-angle.

Formula used.

Perpendicular of tangent through point of contact pass through centre of circle

Sum of all angles of quadrilateral is 360°

Diagonals of square bisect each other at 90°

Solution.

Join AB and OT

As OA and OB are perpendicular to each other

AOB = 90°

If tangents from point A and B are drawn

Then;

OAT = OBT = 90°

Sum of all angles of quadrilateral is 360°

OAT + OBT + AOB + ATB = 360°

90° + 90° + 90° + ATB = 360°

ATB = 360° - 270°

= 90°

All angles of quadrilateral are 90°

Hence quadrilateral can be either square or rectangle

OA = OB (Both are radius of same circle)

Then given quadrilateral is a square

In square diagonals are equal

Hence AB = OT

And diagonal bisect each other at 90°

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