Question

ABCD is a trapezium is which $AB\parallel CD$ and $AD=BC$ (see fig ). Show that $\angle A=\angle B$.

Answer 1

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

Construction : Extends $AB$ and draw a line through $C$ point to $DA$ intersecting $AB$ produced at $E$

Prrof: $AD||CE$ (from construction) &

$AE||DC$ (AS $AB||CD$, & $AB$ is extended)

$AECD$ is a parallelogram.

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

$\therefore AD=CE$ (opposite sides of parallelogram are equal)

But $AD=BC$ (Given)

$\Rightarrow BC=CE$

So, $\angle CEB=\angle CBE$ ...(1) ( In $\Delta BCE$, angles opposite to equal sides are equal)

For $AD||CE$,

& $AE$ is the transversal,

$\angle A+\angle CEB={180}^o$ [interior angles on same side of transversal is supplementary]

$\angle A={180}^o-\angle CEB$ ....(2)

Also $AE$ is line,

so, $\angle B+\angle CE={180}^o$ (liner pairs)

$\angle B+\angle CBE={180}^0$ (from (1))

$\angle B={180}^o-\angle CBE$ ....(3)

from (2) and (3)

$\angle A=\angle B$

$\therefore$ The answer is $\angle A=\angle B$