Question

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

Answer 1

From the figure,

$\Rightarrow \angle ADE+\angle ADC={180}^o$ (Linear pair) ....... (i)

Also, $\square ABCE$ is a cyclic quadrilateral.

So, $\angle AED+\angle ABC={180}^o$ ....... (ii) (opposite angles of cyclic quadrilateral sum upto ${180}^o$)

$\angle ADC=\angle ABC$ ....... (iii) (Opposite angles of a parallellogram are equal)

Using (i), (ii) and (iii),

$\angle AED+\angle ABC=\angle ADE+\angle ABC$

$\Rightarrow \angle AED=\angle ADE$

Now, In $△AED$,

$\angle AED=\angle ADE$

$\Rightarrow AD=DE$ (Sides opposite to equal angles)