Question

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

Answer 1

Given ABCD is a parallelogram passing through A, B, C and intersecting CD at E.
$\angle B=\angle D$ [Opposite angles of a parallelogam are equal] $\to \left(1\right)$
Since ABCE is a cyclic quadrilateral.
Recall that "exterior angle of a cyclic quadrilateral is equal to interior opposite angle".
Hence $\angle B=\angle AED\to \left(2\right)$
From (1) and (2), we get
$\angle D=\angle AED$
Therefore AE = AD [Since in a triangle, sides opposite to equal angles are equal]