If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection,prove that the chords are equal.

(N/A) Given that $AB$ and $CD$ are two chords of a circle with centre $O$,intersecting at a point $E$. $PQ$ is a diameter passing through $E$,such that $\angle AEQ = \angle DEQ$. We need to prove that $AB = CD$.
Draw perpendiculars $OL$ and $OM$ on chords $AB$ and $CD$,respectively.
In $\Delta OLE$,$\angle OLE = 90^{\circ}$. Therefore,$\angle LOE = 180^{\circ} - 90^{\circ} - \angle LEO = 90^{\circ} - \angle LEO$.
Since $\angle LEO = \angle AEQ$ (vertically opposite angles),we have $\angle LOE = 90^{\circ} - \angle AEQ$.
Similarly,in $\Delta OME$,$\angle OME = 90^{\circ}$. Therefore,$\angle MOE = 90^{\circ} - \angle MEO = 90^{\circ} - \angle DEQ$.
Since it is given that $\angle AEQ = \angle DEQ$,it follows that $\angle LOE = \angle MOE$.
Now,in triangles $OLE$ and $OME$:
$1$. $\angle LEO = \angle MEO$ (Given $\angle AEQ = \angle DEQ$ and vertically opposite angles)
$2$. $\angle LOE = \angle MOE$ (Proved above)
$3$. $EO = EO$ (Common side)
By $ASA$ congruence criterion,$\Delta OLE \cong \Delta OME$.
By $CPCT$,$OL = OM$.
Since chords equidistant from the centre of a circle are equal in length,it follows that $AB = CD$.