In a cyclic quadrilateral $ABCD$, if $AB \| CD$ and $\angle B = 70^o$, find the remaining angles.
Given:
In a cyclic quadrilateral $ABCD$, $AB \| CD$ and $\angle B = 70^o$.
To do:
We have to find the remaining angles.
Solution:
$ABCD$ is a cyclic quadrilateral.
$\angle B + \angle D = 180^o$
$70^o +\angle D = 180^o$
$\angle D = 180^o-70^o = 110^o$
$AB \| CD$
This implies,
$\angle A + \angle D = 180^o$ (Sum of cointerior angles)
$\angle A+ 110^o= 180^o$
$\angle A= 180^o- 110^o = 70^o$
Similarly,
$\angle B + \angle C = 180^o$
$70^o + \angle C = 180^o$
$\angle C = 180^o - 70^o = 110^o$
Hence, $\angle A = 70^o, \angle C = 110^o$ and $\angle D = 110^o$.
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