Question

a trapezium will be cyclic if non parallel sides are equal .prove it

Pls answer quiuk

Answer 1

Answer.

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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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