Question

If the non- parallel side of a trapezium are equal, prove that is cyclic ​

Answer 1

Answer:

Step-by-step explanation:

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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)

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Answer 2

In ΔAED and ΔBFC,

AD = BC (Given)

DE = CF (Distance between parallel sides is same)

∠AED = ∠BFC = 90°

ΔAED ≅ ΔBFC (RHS Congruence criterion) Hence ∠DAE = ∠CBF (CPCT) … (1) Since AB||CD, AD is transversal ∠DAE + ∠ADC = 180° (Sum of adjacent interior angles is supplementary) ⇒ ∠CBF + ∠ADC = 180° [from (1)] Since sum of opposite angles is supplementary in trapezium ABCD. Thus ABCD is a cyclic trapezium