Question

$ABCD$ is a trapezium in which $AB||CD$ and $AD=BC$, then show that $\angle C=\angle D.$

Answer 1

Let the line $AB$ is extended up to point E such that $AD||EC$

& $AE||DC$

in $AECD$ both pair of opposite sides parallel

$AECD$ is a parallelogram

$AD=CE$ (opposite side of parallelogram)

$AD=BC$

$\Rightarrow BC=CE$

$\angle CEB=\angle CBE$

$AD||CE$

$AE$ is the transversal

$\angle A+\angle CEB={180}^o$

$\angle A={180}^o-\angle CEB$

$\left(2\right)$ and $\left(3\right)$

$\angle A=\angle B$

Similarly $\angle C=\angle D$

Also $AE$ is a line

$\angle B+\angle CBE={180}^o$

$\angle B+\angle CEB={180}^o$

$\angle B={180}^o-\angle CEB$