if two non parallel sides of a trapezium are equal prove that it is cyclic

**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)

Cheers!

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