# 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.