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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Updated on: 10-Oct-2022

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