Question

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

Answer 1

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

Answer 2

Hello mate ☺

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Solution:

It is given that ABCD is a trapezium with AB∥CD and AD=BC

We need to prove that ABCD is a cyclic quadrilateral.

Construction: Draw AM⊥CD and BN⊥CD

In ∆AMD and ∆BNC, we have

AD=BC            (Given)

∠AMD=∠BNC          (Each equal to 90°)

AM=BN        (Distance between two parallel lines is constant.)

Therefore, by RHS congruence rule, we have ∆AMD≅∆BNC

⇒∠D=∠C        (Corresponding parts of congruent triangles are equal)   ........ (1)

We also have ∠A+∠D=180′      (Co-interior angles, AB∥CD)     ......... (2)

From (1) and (2), we can say that ∠A+∠C=180°

⇒ ABCD is a cyclic quadrilateral.

(If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.)

I hope, this will help you.☺

Thank you______❤

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