If two equal chords of a circle intersect within the circle,prove that the segments of one chord are equal to corresponding segments of the other chord.

(N/A) Let there be a circle with center $O$. Let $AB$ and $CD$ be two equal chords that intersect at a point $E$ inside the circle.
To prove: $AE = DE$ and $CE = BE$.
Construction: Draw $OM \perp AB$ and $ON \perp CD$. Join $OE$.
Proof:
$1$. Since $AB = CD$ (given),the chords are equidistant from the center. Therefore,$OM = ON$.
$2$. In $\Delta OME$ and $\Delta ONE$:
- $OM = ON$ (Proved above)
- $OE = OE$ (Common side)
- $\angle OME = \angle ONE = 90^\circ$ (By construction)
By $RHS$ congruence criterion,$\Delta OME \cong \Delta ONE$.
$3$. By $CPCT$,$ME = NE$ (Equation $1$).
$4$. Since $OM \perp AB$,$M$ is the midpoint of $AB$,so $AM = MB = \frac{1}{2} AB$.
$5$. Since $ON \perp CD$,$N$ is the midpoint of $CD$,so $CN = ND = \frac{1}{2} CD$.
$6$. Since $AB = CD$,then $\frac{1}{2} AB = \frac{1}{2} CD$,which implies $AM = ND$ (Equation $2$).
$7$. Adding equations $1$ and $2$: $AM + ME = ND + NE$,which gives $AE = DE$.
$8$. Subtracting $AE = DE$ from $AB = CD$,we get $AB - AE = CD - DE$,which implies $BE = CE$.
Thus,the segments of one chord are equal to the corresponding segments of the other chord.