Q. 44.3( 3 Votes )

# 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.

Answer :

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

In quadrilateral OATB

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)

If adjacent sides are equal

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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