In a cyclic quadrilateral `A B C D ,` if `/_A-\ /_C=60^0,` prove that the smaller of two is `60^0`

Explanation

Here, `ABCD` is a cyclic quadrilateral.
We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. `180^0`
Then,`/_A+/_C=180^0`
But given,`/_A−/_C=60^0`
Adding the two equations,`2/_A=240^^0`
`/_A=120^0`
Therefore,`/_A+/_C=180^0`
`120^0+/_C=180^0`
`/_C=60^0`
The smaller one is `60^0`
.