To prove that "a straight line drawn from the centre of a circle to bisect a chord which is not a diameter is perpendicular to the chord"

Before going to the topic remember these terms,
Chord:-
A line segment connecting two points on a curve. 
Diameter:-
A straight line going through the center of a circle connecting two points on the circumference.
Perpendicular Line:-
Lines that are at right angles (90°) to each other.

When a perpendicular line is drawn from the center of the circle to the chord. Then the perpendicular line bisects (divide into 2 halves) the chord.

Lets take both of them (AOP & BOP) as right angle triangle and we can see that from point O the radius is same. Here the distance of OP is same in both of the triangles. On the other hand, the angle AOP and BOP are also same. Hence, by using the property of S.A.S the length AP and BP are same and thus, are divided equally (bisect) by the chord.
For its derivation lets proceed forward,

Given a circle with centre O and AC = BC
To Prove: OC ⊥ AB
Construction: Join points O, A and O, B.
Proof:
In Δ’s OCA and OCB
OA = OB (Radii of the circle)
OC = OC (Common side)
AC = BC (Given)
Hence ΔOCA ≅ ΔOCB (SSS Congruence rule)
∴ ∠OCA = ∠OCB (CPCT)
From the figure, ∠OCA + ∠OCB = 180° (Linear pair)
∠OCA + ∠OCA = 180°
2∠OCA = 180°
∴ ∠OCA = 90°
Thus OC ⊥ AB
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