If two equal chords of a circle intersect,prove that the parts of one chord are separately equal to the parts of the other chord.

(N/A) $AB$ and $CD$ are two equal chords of a circle with centre $O,$ which intersect each other at $M.$ We have to prove that:
$(i)$ $MB = MC$ and
$(ii)$ $AM = MD$
Draw $OE \perp AB$ and $OF \perp CD$ from the centre $O.$
Since the perpendicular from the centre to a chord bisects the chord,we have:
$AE = \frac{1}{2} AB$ and $FD = \frac{1}{2} CD$
Since $AB = CD,$ it follows that $\frac{1}{2} AB = \frac{1}{2} CD,$ so $AE = FD$ $...(1)$
Since equal chords are equidistant from the centre,$OE = OF.$
Now,in $\Delta MOE$ and $\Delta MOF$:
$OE = OF$ [Proved above]
$OM = OM$ [Common side]
$\angle OEM = \angle OFM = 90^\circ$ [By construction]
Therefore,$\Delta MOE \cong \Delta MOF$ [By $RHS$ congruence rule]
Thus,$ME = MF$ $...(2)$
Subtracting $(2)$ from $(1),$ we get:
$AE - ME = FD - MF$
$\Rightarrow AM = MD$ [Proved part $(ii)$]
Again,$AB = CD$ and $AM = MD.$
Subtracting these,we get $AB - AM = CD - MD.$
Therefore,$MB = MC$ [Proved part $(i)$]