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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