Question

AB is the diameter of a circle with centre ‘O’. A line MN touches the given circle at point R and cuts the tangents to the circle through A and B at M and N, respectively. Prove that: ∠MON = 90°

Answer 1

Given

AB is the diameter of the circle with centre O.

MN is a tangent touching the circle at R.

Tangent at A meets MN at M.

Tangent at B meets MN at N.

OA and OB are radii.

∠MON = 90°

Step 1: Use tangent–radius properties

Since MN touches the circle at R,

OR ⊥ MN      ...[1]

Since AM is tangent at A,

OA ⊥ AM     ...[2]

Since BN is tangent at B,

OB ⊥ BN      ...[3]

Step 2: AB is the diameter, so

OA and OB lie on the same straight line.

A line perpendicular to OA is also perpendicular to OB.

From (2) and (3):

AM ∥ BN

Step 3: Use transversal MN

MN intersects AM at M and BN at N.

So, ∠OMN and ∠MNO form the two acute angles between a pair of parallel lines AM and BN and the transversal MN.

The angle between OM and ON is an exterior angle of the “triangle made by two perpendicular lines and a transversal”, so:

∠MON = 180° − (∠OMA + ∠ONB)

∠OMA = 90°, ∠ONB = 90°

∠MON = 180° − (90° + 90°)

∠MON = 90°