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Question

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

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Solution

Given: ABCD is a trapezium where AB||CD and AD = BC

To prove: ABCD is cyclic.
Construction: Draw DL⊥AB and CM⊥AB.

Proof: In ΔALD and ΔBMC,
AD = BC (given)
DL = CM (distance between parallel sides)
∠ALD = ∠BMC (90°)
ΔALD ≅ ΔBMC (RHS congruence criterion)
⇒ ∠DAL = ∠CBM (C.P.C.T) (1)
Since AB||CD,
∠DAL + ∠ADC = 180° (sum of adjacent interior angles is supplementary)
⇒ ∠CBM + ∠ADC = 180° (from (1))
⇒ ABCD is a cyclic trapezium (Sum of opposite angles is supplementary)

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