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

Question

In a cyclic quadrilateral ABCD, if &#x2220; A - &#x2220; C =60&#xB0;, prove that the smaller of two is 60&#xB0;.

Philoid · Accepted Answer

WE have,

∠A - ∠C = 60^o (i)

Since, ABCD is a cyclic quadrilateral

Then,

∠A + ∠C = 180^o (ii)

Adding (i) and (ii), we get

∠A - ∠C + ∠A + ∠C = 60^o + 180^o

2 ∠A = 240^o

∠A = 120^o

Put value of ∠A in (ii), we get

120^o + ∠C = 180^o

∠C = 60^o