Question
$ABCD$ is a cyclic qudrilateral in which: $ \angle\text{DBC}=80^\circ$ and $\angle\text{BAC}=40^\circ.$ Find $\angle\text{BCD}.$
✓
Answer
$\angle\text{BAC}=\angle\text{BDC}=40^\circ$ [Angle in same segment]
In by angle sum property $\angle\text{DBC}+\angle\text{BCD}+\angle\text{BDC}=180^\circ$
$\Rightarrow80^\circ+\angle\text{BCD}+40^\circ=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-80^\circ-40^\circ=60^\circ$
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