Workings

$\triangle ADE \cong \triangle CFE \left(AAS\right)$

$\angle AED = \angle CEF $( vertically opposite angles )

$\angle CFE= \angle EDA $( alternate angles )

$AE=EC $( E midpoint )

$ii.$ADCF is a parallelogram because diagonals bisect each other.

Where is help needed

How should the fact that area of $\triangle ABC$ is equal to parallelogram ADCF be proved ?

Show that if $DE=AE$, then $\angle ADC=90^{\circ}$