$ABCD$ is a trapezium in which $AB \parallel CD$ and $AD = BC$ (see Fig). Show that diagonal $AC =$ diagonal $BD$.

(A) Given: $ABCD$ is a trapezium with $AB \parallel CD$ and $AD = BC$.
Construction: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E$.
Proof:
$1$. Since $AD \parallel CE$ (by construction) and $AE \parallel DC$ (given $AB \parallel DC$),$AECD$ is a parallelogram.
$2$. Therefore,$AD = CE$ (opposite sides of a parallelogram). Since $AD = BC$ (given),we have $BC = CE$.
$3$. In $\Delta BCE$,$BC = CE$,so $\angle CBE = \angle CEB$ (angles opposite to equal sides).
$4$. Also,$\angle ABC + \angle CBE = 180^\circ$ (linear pair). Since $\angle CEB = \angle CBE$,we have $\angle ABC + \angle CEB = 180^\circ$.
$5$. In $\Delta ABC$ and $\Delta BAD$:
- $AB = BA$ (common side)
- $AD = BC$ (given)
- $\angle DAB = \angle CBA$ (since $\angle D + \angle A = 180^\circ$ and $\angle C + \angle B = 180^\circ$,and $AD=BC$ implies an isosceles trapezium).
$6$. By $SAS$ congruence rule,$\Delta ABC \cong \Delta BAD$.
$7$. Therefore,$AC = BD$ (by $CPCT$).