Let BD bisect $\angle ABC$ in $\Delta ABC$. Given $\angle ABC$ = 60 degrees and AB = BC + CD, find $\angle BAC$

Question

I tried using angle bisector theorem and cosine rule but to no avail. Please help.

Answer 1

Construct the point $E$ on $AB$ such that $EB = BC$. Then $AE = BA - BE = BA - BC = CD$.

Besides $\triangle BED \cong \triangle BCD$ by SAS. Hence $ED = CD = AE$.

We have $\angle EAD = \angle EDA = \alpha$. $\angle EDB = \angle CDB = \frac12(180^\circ - \alpha)$.

We also have $\angle BED = 2\alpha$. Now considering the interior angles of $\triangle BED$,

$$30^\circ + 2\alpha + \frac12(180^\circ - \alpha) = 180^\circ$$