Question

$ABCD$ is a trapezium in which $AB\parallel CD$ and $AD=BC$ (See figure). Show that
(i) $\angle A=\angle B$
(ii) $\angle C=\angle D$
(iii) $△ABC\cong \Delta BAD$

Answer 1

Construction :

Draw a line through $C$ parallel $DA$

Intersecting $AB$ produced $E$.

Now,

$\left(i\right)$

$AB\parallel CD$ (given)

$AD\parallel EC$ (By construction)

So,

$ADCE$ is a parallelogram

Hence,

$CE=AD$ (Opposite sides of parallelogram)

$AD=BC$ (Given)

We know that.

$\angle A+\angle E={180}^{\circ }$ (interior angles on the same side of the transverse $AE$)

$\angle E={180}^{\circ }-\angle A$

Also, $BC=CE$

$\angle E=\angle CBE={180}^{\circ }-\angle A$

$\angle ABC={180}^{\circ }-\angle CBE$ (Since, $ABE$ is straight line)

$\angle ABC={180}^{\circ }-({180}^{\circ }-\angle A)$

$\angle ABC={180}^{\circ }-{180}^{\circ }+\angle A$

Therefore,

$\angle B=\angle A$ ---- $\left(1\right)$

$\left(ii\right)$

$\angle A+\angle D=\angle B+\angle C={180}^{\circ }$ (Angles at same side of transverse)

$\angle A+\angle D=\angle B+\angle C$

$\angle D=\angle C$ ($\angle B=\angle A$ from $\left(1\right)$)

$\left(iii\right)$

In $\Delta ABC$ and $\Delta BAD$

$AB=AB$ (common)

$\angle DBA=\angle CBA$ (from $\left(1\right)$)

$AD=BC$ (Given)

$\Delta ABC\cong \Delta BAD$ (By SAS congruence rule)