Question

If the non-parallel sides of trapezium are equal, prove that it is cyclic.

Answer 1

In $△AED$ and $△BFC$,

$AD=BC$ (Given)

$DE=CF$ (Distance between parallel sides is same)

$\angle AED=\angle BFC={90}^{\circ }$

$△AED\cong △BFC$ (RHS Congruence criterion)

Hence $\angle DAE=\angle CBF$ (CPCT) … $\left(1\right)$

Since $AB$ is parallel $CD,AD$ is transversal

$\angle DAE+\angle ADC={180}^{\circ }$ (Sum of adjacent interior angles is supplementary)

$\Rightarrow \angle CBF+\angle ADC={180}^{\circ }$ [from $\left(1\right)$]

Since sum of opposite angles is supplementary in trapezium $ABCD$.

Thus $ABCD$ is a cyclic trapezium