Question

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

Answer 1

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)