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.