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)



  • 312

sorry forgot the fig.

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