In triangle $ABC$ find angle $\angle BAC$ given that... 
In triangle $ABC$, $D$ is a point on $BC$ such that $CD=2BD$.Also we know that $\angle B=45$ and $\angle ADC=60$.Find angle $\angle BAC$.

I tried to find two similar triangles but failed! I heared the aimed solution method uses similar triangles...



Let $E$ be the projection of $A$ on $BC$ and let us assume $AE=\sqrt{3}$.
We have $ED=1$ and $EB=\sqrt{3}$, from which $DB=\sqrt{3}-1$ and $CE=2\sqrt{3}-3$.
By the Pythagorean theorem, $AB=\sqrt{6}$ and $AC=2\sqrt{6-3\sqrt{3}}=\sqrt{6}(\sqrt{3}-1)$.
By the cosine theorem:
$$\cos\widehat{BAC}=\frac{AB^2+AC^2-BC^2}{2\,AB\cdot AC} = \frac{6+12(2-\sqrt{3})-18(2-\sqrt{3})}{12(\sqrt{3}-1)}=\frac{1}{2}$$
hence $\widehat{BAC}=\color{red}{60^\circ}$ as wanted.
