Question

State True or False

$ABCD$ is trapezium in which $AB||CD$ and $AD||BC$, then BD=AC

• A. True
• B. False

Answer 1

The correct option is A True

Given:$ABCD$ is a trapezium where $AB\parallel CD$ and $AD=BC$

From the figure above in the question,$AB\parallel CE$

and $AE\parallel DC$ as $AB\parallel CD$ and $AB$ is extended.

In $AECD$, both pair of opposite sides are parallel,

$AECD$ is a parallelogram.

$\therefore AD=CE$ since opposite sides of parallelogram are equal.

But $AD=BC$(given)

$\Rightarrow BC=CE$

So,$\angle CEB=\angle CBE$ ..........$\left(1\right)$ since in $△BCE,$ angles opposite to equal sides are equal.

For $AD\parallel CE,$ and $AE$ is the transversal,

$\angle A+\angle CEB={180}^{\circ }$(interior angle on the same side of transversal is supplementary)

$\angle A={180}^{\circ }-\angle CEB$ .........$\left(2\right)$

Also,$AE$ is a line

So,$\angle B+\angle CBE={180}^{\circ }$(linear pair)

$\angle B={180}^{\circ }-\angle CEB$ .........$\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$ we have

$\angle A=\angle B$

In $△ABC$ and $△BAD,$

$AB=BA$(common)

$\angle B=\angle A$(proved above)

$BC=AD$(Given)

$\therefore △ABC\cong △BAD$(SAS congruence rule)

$\therefore AC=BD$ by CPCT