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  • If non-parallel sides of a trapezium are equal, prove that it is cyclic.

If non-parallel sides of a trapezium are equal, prove that it is cyclic.

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Given: non-parallel sides of a trapezium are equal

Let ABCD is a trapezium where AB || CD & non-parallel sides are equal

i.e., AD = BC

To Prove: ABCD is a cyclic quadrilateral

Proof:

Draw DE \perp AB

And CF \perp AB

In \triangleADE & \triangleBCF

\angleAED = \angleBFC  (Both are 90°)

AD = BC (Given)

DE = CF (perpendicular distance between parallel sides will remain the same)

\because \triangleADE \cong \triangleBCF  (RHS congruence rule)

So\angleDAE = \angleCBF    (CPCT)

i.e. \angleA = \angleB ....…(i)

Now, AB || DC, AD is a transversal line

\angleA + \angleD = 180°        [Interior angles on the same side of the transversal are supplementary]

\angleB + \angleD = 180°         [From (i)]

So, in ABCD, the sum of one pair of opposite angle is 180°.

Therefore, ABCD is a cyclic quadrilateral

Hence Proved.

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infoexpert24

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