If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides,prove that the quadrilateral so formed is cyclic.

(N/A) Let $\Delta ABC$ be an isosceles triangle with $AB = AC$. $A$ line $DE$ is drawn parallel to $BC$ such that $D$ lies on $AB$ and $E$ lies on $AC$.
We need to prove that the quadrilateral $BCED$ is a cyclic quadrilateral.
In $\Delta ABC$,since $AB = AC$,the angles opposite to equal sides are equal,so $\angle ABC = \angle ACB$ (or $\angle 1 = \angle 2$ as per the figure).
Since $DE \parallel BC$,the interior angles on the same side of the transversal $AB$ are supplementary.
Therefore,$\angle BDE + \angle ABC = 180^{\circ}$ (or $\angle 3 + \angle 1 = 180^{\circ}$).
Substituting $\angle 1 = \angle 2$,we get $\angle 3 + \angle 2 = 180^{\circ}$.
This means the sum of the opposite angles of the quadrilateral $BCED$ is $180^{\circ}$.
Since the sum of a pair of opposite angles of the quadrilateral $BCED$ is $180^{\circ}$,the quadrilateral $BCED$ is cyclic.