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

Given:
Non-parallel sides PS and QR of a trapezium PQRS are equal.

Construction:
Draw \(SM\perp{PQ}\) and \(RN\perp{PQ}\).



To prove: ABCD is cyclic trapezium.

Proof:
In \(\triangle{SMP}\) and \(\triangle{RNQ}\), we have,

SP = RQ ...(Given)
SM = RN
(Distance between two parallel twolines is always equal)

and \(\angle{SMP}\) = \(\angle{RNQ}\)
(Each \(90^\circ\))
By RHS criterion, we get,

\(\triangle{SMP}\) \(\displaystyle \cong\) \(\triangle{RNQ}\)

Hence, \(\angle{P}\) = \(\angle{Q}\) ...(By CPCT)

Also, \(\angle{PSM}\) = \(\angle{QRN}\)

Now, adding \(90^\circ\) on both sides, we get,

\(\therefore \) \(\angle{PSM}\) + \(90^\circ\) = \(\angle{QRN}\) + \(90^\circ\)
\(\therefore \) \(\angle{MSR}\) + \(\angle{PSM}\) = \(\angle{NRS}\) + \(\angle{QRN}\)
(Since, \(\angle{MSR}\) = \(\angle{NRS}\) = \(90^\circ\))

\(\therefore \) \(\angle{PSR}\) = \(\angle{QRS}\)
\(\Rightarrow \) \(\angle{S}\) = \(\angle{R}\)
\(\therefore \) \(\angle{P}\) = \(\angle{Q}\) and \(\angle{S}\) = \(\angle{R}\)

\(\because \) Sum of the angles of a quadrilateral is \(360^\circ\)
We get,

\(\angle{P}\) + \(\angle{Q}\) + \(\angle{S}\) + \(\angle{R}\) = \(360^\circ\)

From eq. (i), we get,

2[\(\angle{Q}\) + \(\angle{S}\)] = \(360^\circ\)
\(\therefore \) [\(\angle{Q}\) + \(\angle{S}\)] = \(180^\circ\)

Hence, it is proved that PQRS is a cyclic trapezium.