Question

In a cyclic quadrilateral ABCD, if $\angle A-\angle C={60}^{\circ }$, prove that the smaller of two is ${60}^{\circ }$.

Answer 1

Here, $ABCD$ is a cyclic quadrilateral.

We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. ${180}^o$.

Then, $\angle A+\angle C={180}^o$.

But given, $\angle A-\angle C={60}^{\circ }$.

Adding the two equations,

$⟹$ $2\angle A={240}^{\circ }$

$⟹$ $\angle A={120}^{\circ }$.

Therefore, $\angle A+\angle C={180}^o$

$⟹$ ${120}^o+\angle C={180}^o$

$⟹$ $\angle C={60}^{\circ }$.

$\therefore$ The smaller one is ${60}^{\circ }$.

Hence, proved.