In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°

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Solution

It is given that ∠A – ∠C = 60° and ABCD is a cyclic quadrilateral.

We have to prove that smaller of two is 60°

Since ABCD is a cyclic quadrilateral

So ∠A + ∠C = 180° (Sum of opposite pair of angles of cyclic quadrilateral is 180°) ..… (1)

And,

∠A – ∠C = 60° (Given) ..… (2)

Adding equation (1) and (2) we have

So, ∠C = 60°

Hence, smaller of two is 60°.

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∠ A - ∠ C = 60^{\circ}

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In a cyclic quadrilateral ABCD, if $(∠ B - ∠ D) = 60^{\circ}$ , show that the smaller of the two is $60^{\circ}$ .

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