# Let BD bisect angle ABC in triangle ABC. Given angle ABC = 60 degrees and AB = BC + CD, find angle BAC.

Let BD bisect $\mathrm{\angle }ABC$ in $\mathrm{\Delta }ABC$. Given $\mathrm{\angle }ABC=60$ degrees and $AB=BC+CD$, find $\mathrm{\angle }BAC$
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Step 1

Construct the point E on AB such that $EB=BC$. Then $AE=BA-BE=BA-BC=CD$.
Besides $\mathrm{△}BED\cong \mathrm{△}BCD$ by SAS. Hence $ED=CD=AE$.
Step 2
We have $\mathrm{\angle }EAD=\mathrm{\angle }EDA=\alpha$. $\mathrm{\angle }EDB=\mathrm{\angle }CDB=\frac{1}{2}\left({180}^{\circ }-\alpha \right)$.
We also have $\mathrm{\angle }BED=2\alpha$. Now considering the interior angles of $\mathrm{△}BED$, ${30}^{\circ }+2\alpha +\frac{1}{2}\left({180}^{\circ }-\alpha \right)={180}^{\circ }$.