Question

if the non parallel sides of a Trapezium are equal then prove that it is a cyclic quadrilateral​

Answer 1

Answer

for a quadrilateral to be cyclic the sum of opposite angles in the quadrilateral is 180°

Step-by-step explanation:

TO PROVE ∠A+∠C=180 OR ∠B+∠D=180

DRAW DE AND CF PERPENDICULAR TO AB

IN TRIANGLE aed and bfc we have

ad=bc

∠aed=∠bfc=90°

de=fc (height between parallel sides)

so Δaed  ≅ Δbfc

∠a=∠b(cpct)......................(1)

∠a+∠d=180° (angles on the same side of the transversal are supplementary)

∠b+∠d=180° from eq(1)

∠a+∠b+∠c+∠d=360°

∠a+∠c=180°  from eq(1)

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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