Question

If the non-parallel sides of a trapezium are equal show that it is cyclic

Answer 1

Let the trapezium ABCDABCD be isosceles and the parallel sides be ABAB and CDCD. The equal length non-parallel sides are BCBC, DADA. The angles ∠DAB∠DAB and ∠CDA∠CDA are supplementary in the parallel lines so ∠DAB+∠CDA=π∠DAB+∠CDA=π.

The trapezium is symmetric so the angles ∠DAB∠DAB and ∠ABC∠ABC are equal.

So we have ∠ABC+∠CDA=π∠ABC+∠CDA=π, i.e. we have opposite angles in the trapezium add up to ππ. This is necessary and sufficient for a quadrilateral to be cyclic.

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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