[ThE-dArK-lOrD]

For second problem of geometry, it can be solved without law of sines
Let $O$ be the circumcenter of ABC and let $\omega$ be the circle centered at $D$ with radius $BD$
Let lines $AB,AC$ meet $\omega$ at $P,Q$, respectively. Since $\angle{PBQ}=\angle{BQC}+\angle{BAC}=\frac{1}{2} \cdot (\angle{BDC} + \angle{DOC}) = 90^{\circ}$
We see that $PQ$ is a diameter of $\omega$ and passes through $D$.
Since $\angle{ABC} = \angle{AQP}$ and $\angle{ACB} = \angle{APQ}$,
So $\triangle{ABC} \sim \triangle{AQP}$ and since $M$ is midpoint of $BC$ and $D$ is midpoint of $PQ$
The similarity implies that $\angle{BAM} = \angle{QAD}$ and complete the prove