Question

State true or false:

The following figure shows a trapezium $ABCD$ in which $AB$ is parallel to $DC$ and $AD=BC$, then $AC=BD$

• A. True
• B. False

Answer 1

The correct option is A True

Extend CD to E and join BE such that $BE\parallel AD$

Now, $AD\parallel BE$ and $AB\parallel DE$ $[\because AB\parallel CD$ & CD extended to E]

Therefore ABED is a $\parallel gm$

$\therefore AD=BE$ , but $AD=BC$ , so that BC=BE, then $\angle BEC=\angle BCE$ [angles opposite to equal sides are equal] $\to \left(1\right)$

$AD\parallel BE$ , then $\angle D+\angle BEC={180}^{\circ }$ [Interior angles on same side of traversal are supplementary]

$\angle D={180}^{\circ }-\angle BEC\to \left(2\right)$

& $\angle C+\angle BEC={180}^{\circ }$ [Linear pair]

$\angle C={180}^{\circ }-\angle BEC\to \left(3\right)$

From (1) , (2) & (3), we get

$\angle D=\angle C\to \left(4\right)$

Now in $△ADC$ and $△BCD$,

$AD=BC$ (Given)

$\angle D=\angle C$ (From (4))

$CD=DC$ (Common)

$△ADC\cong △BCD$ [By SAS]

Hence, $AC=BD$ [By CPCT]