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

Given:

In a cyclic quadrilateral $ABCD$, $\angle A - \angle C = 60^o$

To do:

We have to prove that the smaller of two is $60^o$.

Solution:

$\angle A - \angle C = 60^o$...........(i)

$ABCD$ is a cyclic quadrilateral.

This implies,

$\angle A + \angle C = 180^o$...........(ii)                (Sum of the opposite angles)

Adding (i) and (ii), we get,

$2\angle A = 240^o$

$\angle A = 120^o$

Substituting $\angle A = 120^o$ in (i), we get,

$120^o-\angle C = 60^o$

$\angle C = 120^o-60^o$

$\angle C = 60^o$

Therefore, the smaller angle of the two angles is $60^o$.

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Updated on: 10-Oct-2022

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