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If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.

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Given: A pair of opposite sides of a cyclic quadrilateral are equal.

Let ABCD be a cyclic quadrilateral with AD = BC

To Prove: Diagonals are equal, i.e., AC = BD

Proof:

We know that angles in the same segment are equal.

Consider segment CD, \angleDAC = \angleDBC                   

So we can write,  \angleDAO = \angleCBO …(i)

Consider segment AB, \angleADB = \angleACD

So we can write,  \angleADO = \angleBCO…(ii)

In \triangleAOD & \triangleBOC,

\angleDAO = \angleCBO  (from (i))

\angleADO = \angleBCO   (from (ii))

\angleAOD = \angleBOC (vertically opposite angles)

\triangleAOD \cong \triangleBOC  [AAA congruence rule]

AO = BO (CPCT)…(iii)

DO = CO (CPCT)…(iv)

Adding (iii) and (iv),

AO + OC = BO + OD

Hence, AC = BD

Hence proved

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