‘O’ is the centre of the circle.
AB is a chord, M is its midpoint.
Join A, B to ’O’.
Now in ΔOMA and ΔOMB
OA = OB (radii)
OM = OM (common)
MA = MB (given)
∴ ΔOMA s ΔOMB (SSS congruence)
∴ ∠OMA = ∠OMB (C.P.C.T)
But ∠OMA and ∠OMB are linear pair
∴∠OMA = ∠OMB = 90°
i.e., OM ⊥ AB.