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

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#### Solution

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

Draw AM ⊥ CD and BN ⊥ CD.

In ΔAMD and ΔBNC,

AD = BC (Given)

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

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

∴ ΔAMD ≅ ΔBNC (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.

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