$ABCD$ is a cyclic quadrilateral in which $\angle DBC = 80^o$ and $\angle BAC = 40^o$. Find $\angle BCD$.
Given:
$ABCD$ is a cyclic quadrilateral in which $\angle DBC = 80^o$ and $\angle BAC = 40^o$.
To do:
We have to find $\angle BCD$.
Solution:
$ABCD$ is a cyclic quadrilateral.
Join diagonals $AC$ and $BD$.
$\angle DBC = 80^o, \angle BAC = 40^o$
Arc $DC$ subtends $\angle DBC$ and $\angle DAC$ in the same segment
Therefore,
$\angle DBC = \angle DAC = 80^o$
$\angle DAB = \angle DAC + \angle CAB$
$= 80^o + 40^o$
$= 120^o$
$\angle DAC + \angle BCD = 180^o$ (Sum of opposite angles of a cyclic quadrilateral is $180^o$)
$120^o +\angle BCD = 180^o$
$\angle BCD = 180^o- 120^o$
$= 60^o$
Hence $\angle BCD = 60^o$.
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