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If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
The non-parallel sides of a trapezium are equal.
We have to prove that it is cyclic.
Let $PQRS$ be a trapezium in which $PQ \| RS$ and $PS=QR$
Draw $PM \perp RS$ and $QN \perp RS$
In $\triangle PSM$ and $\triangle QRN$,
$\angle PMS=\angle QNR=90^o$
$PM=QN$ (Perpendicular distance between two parallel lines is equal)
Therefore, by RHS congruency,
$\triangle PMS \cong QNR$
$\angle PSR=\angle QRS$.........(i) (CPCT)
$\angle QPS$ and $\angle PSR$ are on the same side of the transversal $PS$
$\angle QPS+\angle PSR=180^o$
$\angle QPS+\angle QRS=180^o$ [From (i)]
Opposite angles are supplementary.
Hence, the given trapezium is cyclic.
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