Question

ABCD is a trapezium where $AB\parallel CD$ and AD = BC.Prove that ABCD is cyclic.

Answer 1

Given: ABCD is a trapezium where $AB\parallel CD$ and AD = BC
To prove: ABCD is cyclic
Construction: Draw $DL\perp AB$ and $CM\perp AB$

Proof: In $\Delta ALD$ and $\Delta BMC$,
AD = BC (given)
DL = CM (distance between parallel sides)
$\angle ALC=\angle BMC\left({90}^{\circ }\right)$
$\Delta ALD\cong \Delta BMC$ (RHS congruence criterion)
$\Rightarrow \angle DAL+\angle ADC={180}^{\circ }$ (sum of adjacent interior angles is supplementary)
$\Rightarrow \angle CBM+\angle ADC={180}^{\circ }$ (from (1))
$\Rightarrow$ ABCD is a cyclic trapezium (Sum of opposite angles is supplementary)