Question

In a trapezium $ABCD$ with $AB||CD$, it is given that $AD$ is not parallel to $BC$. Is $△ABC\cong △ADC$? Given reasons.

Answer 1

Yes,$△ABC\cong △ADC$

$\because AC=AC$(common)

$\angle CAB=\angle ACD$(alternate angles since $AB\parallel CD$)

$\angle ACB=\angle CAD$ (alternate angles since $AB\parallel CD$)

So by ASA congruency

$△ABC\cong △ADC$

Hence proved.