We have to prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.

Solution:

Let a circle with centre $O$ and a chord $AB$, $M$ be the mid point of $AB$ and $OM$ is joined and produced to meet the minor arc $AB$ at $N$.

Join $OA$ and $OB$

$M$ is the mid point of $AB$

This implies,

$OM \perp AB$

In $\triangle OAM$ and $\triangle OBM$,

$OA = OB$ (Radii of the circle)

$OM = OM$ (common)

$AM = BM$ ($M$ is the mid point of $AB$)

Therefore, by SSS axiom,

$\triangle OAM = \triangle OBM$

This implies,

$\angle AOM = \angle BOM$ (CPCT)

$\angle AOM = \angle BOM$

These are the angles at the centre made by arcs $AN$ and $BN$.