8. 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,

Thus, \(\angle{PSM}\) + \(90^\circ\) = \(\angle{QRN}\) + \(90^\circ\)

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

So, \(\angle{PSR}\) = \(\angle{QRS}\)

i.e., \(\angle{S}\) = \(\angle{R}\)

Thus, \(\angle{P}\) = \(\angle{Q}\) and \(\angle{S}\) = \(\angle{R}\)

Since, 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.