If two equal chords of a circle intersect within the circle,prove that the line joining the point of intersection to the centre makes equal angles with the chords.

(N/A) Let $AB$ and $CD$ be two equal chords of a circle with centre $O$,intersecting at point $E$ inside the circle.
We need to prove that $\angle OEA = \angle OED$.
Draw $OM \perp AB$ and $ON \perp CD$.
In $\Delta OME$ and $\Delta ONE$:
$1$. $OM = ON$ (Equal chords are equidistant from the centre).
$2$. $OE = OE$ (Common side).
$3$. $\angle OME = \angle ONE = 90^\circ$ (By construction).
Therefore,by $RHS$ congruence criterion,$\Delta OME \cong \Delta ONE$.
By $CPCT$,$\angle OEM = \angle OEN$.
Since $M$ lies on $AB$ and $N$ lies on $CD$,this implies $\angle OEA = \angle OED$.