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

Explanation

Two radii OA and OB of a circle with centre at O are prependicular to ecah other. Two tangents ST and RT at the points A and B respectively intersect each other at T.
Let us join the points A, B and O, T. Let AB and OT intersect eacth other at C.
To prove : `AB=OT` and AB and OT bisect each other at a right angle.
Proof : In the quadrilateral `OATB, OA bot OB` (given),` therefore angle AOB=90^(@), OA bot TS, angle OAT=90^(@)and OB bot TR therefore angle OBT =90^(@), i.e., ` three of the angles of OATB are right-angles. So, the fourth angle is also a right angle.
`therefore` each of the angles of OATB is a right angle.
Again, `OA = OB[because` radii of same circle]
`therefore OA = OB = BT=TA.`
`therefoer ` each of the side of the quadrillateral OATB are eqial and each of the angles is a right angle.

`therefore OATB` is a square and AB and OT are two of its diagonals.
We know that diagonals of a square are equal and the diagonals bisect each other at right-angles.
`thereforeAB=OT` and AB and OT bisect each other at a right angle. [Proved]