Question

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

Answer 1

Consider a trapezium ABCD with AB | |CD and BC = AD.

Draw AM ∠CD and BN ∠CD.

In ∆AMD and ∆BNC,

AD = BC (Given)

∠AMD =∠BNC (By construction, each is 90°)

AM = BM (Perpendicular distance between two parallel lines is same)

∠∆AMD ∠∆BNC (RHS congruence rule)

∠ADC = ∠BCD (CPCT) ... (1)

∠BAD and ∠ADC are on the same side of transversal AD.

∠BAD + ∠ADC = 180° ... (2)

∠BAD + ∠BCD = 180° [Using equation (1)]

This equation shows that the opposite angles are supplementary.

Therefore, ABCD is a cyclic quadrilateral.