Question

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

Answer 1

Consider a trapezium ABCD with AB | |CD and BC = AD.
Construction:
Draw $AM\perp CD$ and $BN\perp CD$.

In $\Delta AMDand\Delta BNC$
AD = BC (Given)
$\angle AMD=\angle BNC$ (By construction, each angle is ${90}^{\circ }$)
AM = BN (Perpendicular distance between two parallel lines is same)
$\therefore \Delta AMD\cong \Delta BNC$ (RHS congruence rule)
$\Rightarrow \angle ADC=\angle BCD\left(CPCT\right)...\left(1\right)$

$\angle BADand\angle ADC$ are on the same side of transversal AD.
$\angle BAD+\angle ADC={180}^{\circ }...\left(2\right)$
$\angle BAD+\angle BCD={180}^{\circ }$ [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.