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

Given:ABCDis a trapezium where non-parallel sidesADandBCare equal.Construction:DMandCNare perpendicular drawn onABfromDandC, respectively.To prove:ABCDis cyclic trapezium.Proof:In△DAMand△CBN,AD=BC...[Given]∠AMD=∠BNC...[Right angles]DM=CN...[Distance between the parallel lines]Therefore,△DAM≅△CBNbyRHScongruence condition.Now,∠A=∠B...[by CPCT]Also,∠B+∠C=180°[Sum of the co-interior angles]⇒∠A+∠C=180°.Thus,ABCDis a cyclic quadrilateral as the sum of the pair of opposite angles is180°.