$ABCD$ is a trapezium in which $AB \parallel CD$ and $AD = BC$ (see Fig). Show that $\Delta ABC \cong \Delta BAD$.

(N/A) To prove $\Delta ABC \cong \Delta BAD$:
Construction: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E$.
$1$. Since $AD \parallel CE$ and $AE \parallel DC$,$AECD$ is a parallelogram.
$2$. Therefore,$AD = CE$ (opposite sides of a parallelogram).
$3$. Given $AD = BC$,so $BC = CE$. Thus,$\angle CEB = \angle CBE$ (angles opposite to equal sides in $\Delta BCE$).
$4$. Also,$\angle ABC + \angle CBE = 180^{\circ}$ (linear pair) and $\angle BAD + \angle ADC = 180^{\circ}$ (consecutive interior angles).
$5$. Since $AD \parallel CE$,$\angle ADC + \angle DCE = 180^{\circ}$.
$6$. By comparing these,we can establish $\angle ABC = \angle BAD$.
$7$. In $\Delta ABC$ and $\Delta BAD$:
- $AB = BA$ (Common side)
- $BC = AD$ (Given)
- $\angle ABC = \angle BAD$ (Proved above)
$8$. By $SAS$ congruence rule,$\Delta ABC \cong \Delta BAD$.