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

Hi!
Here is the answer to your question.
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!

  • 52
View Full Answer

ith AB | |CD and BC = AD.

Draw AM ⊥ CD and BN ⊥ CD.

In ΔAMD and ΔBNC,

AD = BC (Given)

∠AMD = ∠BNC (By construction, each is 90°)

AM = BM (Perpendicular distance between two parallel lines is same)

∴ ΔAMD ≅ ΔBNC (RHS congruence rule)

∴ ∠ADC = ∠BCD (CPCT) ... (1)

∠BAD and ∠ADC are on the same side of transversal AD.

∠BAD + ∠ADC = 180° ... (2)

∠BAD + ∠BCD = 180° [Using equation (1)]

This equation shows that the opposite angles are supplementary.

Therefore, ABCD is a cyclic quadrilateral.

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