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.