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)
∠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)