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.
To do:
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 $BN$.
Thetrefore,
$Arc\ AN = Arc\ BN$
Hence, $N$ divides the arc in two equal parts.
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