Question
In the given figure, BD = DC and $\angle\text{CBD}=30^\circ,$ find $\angle\text{BAC}.$
✓
Answer
BD = DC
$\Rightarrow\ \angle\text{BCD}=\angle\text{CBD}=30^\circ$
In $\triangle\text{BCD},$ we have:
$\angle\text{BCD}+\angle\text{CBD}+\angle\text{CDB}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ 30^\circ+30^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\ \angle\text{CDB}=(180^\circ-60^\circ)=120^\circ$
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, $\angle\text{CDB}+\angle\text{BAC}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{BAC}=180^\circ$
$\Rightarrow\ \angle\text{BAC}=(180^\circ-120^\circ)=60^\circ$
$\therefore\ \angle\text{BAC}=60^\circ$
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