Question

if the non parallel sides of a trapezium is equal prove its a cyclic quadrilateral

Answer 1

ABCD is a parallelogram in which AB || CD
And non parallel sides AD and BC are equal.

Draw a line CE parallel to AD
CE || AD and AE || CD
Therefore, AECD is a parallelogram
angle CDA = angle CEA  (opp angles of ||gm)
AD = CE (opp sides of ||gm)
But AD = BC (given)
Therefore,  CE = BC
angle CEB = angle CBE (opp angles of equal sides)
angle CEB + angle CEA = 180⁰


angle CEB + angle CDA = 180⁰

Trapezium is cyclic as the sum of the pair of oppo angles is 180⁰. 

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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