Question

In a cyclic quadrilateral ABCD, if (∠B – ∠D) = 60°, show that the smaller of the two is 60°.

Answer 1

It is given that ABCD is a cyclic quadrilateral

(∠B – ∠D) = 60^o …….. (1)

We know that

(∠B + ∠D) = 180^o ……. (2)

By adding both the equations

∠B – ∠D + ∠B + ∠D = 60^o + 180^o

So we get

2 ∠B = 240^o

By division

∠B = 120^o

By substituting equation it in equation (1)

(∠B – ∠D) = 60^o

120^o – ∠D = 60^o

On further calculation

∠D = 120^o – 60^o

By subtraction

∠D = 60^o

Therefore, the smaller of the two angles ∠D = 60^o.