$ABCD$ is a parallelogram. The circle passing through $A, B$ and $C$ intersects $CD$ (produced if necessary) at $E$. Prove that $AE = AD$.

(N/A) Given: $ABCD$ is a parallelogram. $A$ circle passes through $A, B$ and $C$,intersecting $CD$ at $E$.
Step $1$: Since $ABCE$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle AEC + \angle B = 180^{\circ}$ --- $(1)$
Step $2$: In parallelogram $ABCD$,opposite angles are equal.
Therefore,$\angle D = \angle B$ --- $(2)$
Step $3$: From equations $(1)$ and $(2)$,we get:
$\angle AEC + \angle D = 180^{\circ}$ --- $(3)$
Step $4$: Since $D, E, C$ lie on a straight line,$\angle AEC$ and $\angle AED$ form a linear pair.
Therefore,$\angle AEC + \angle AED = 180^{\circ}$ --- $(4)$
Step $5$: Comparing equations $(3)$ and $(4)$:
$\angle D = \angle AED$
Step $6$: In $\Delta ADE$,since the base angles $\angle D$ and $\angle AED$ are equal,the sides opposite to these angles must be equal.
Therefore,$AE = AD$. Hence proved.