Proving that $\angle B\le60^\circ$ 
$A$ is the smallest angle of $\triangle ABC$. The height from $A$ over $BC$ is equal to the median from $B$. Prove that $\angle B\le 60^\circ$.

I tried extending the median to construct a parallelogram and expressed the area in terms of $BC$, the height over $BC$ and the median, but didn't succeed.

Note that the median partitions the triangle into two triangles with the same area. Hence if $M$ is the midpoint of $AC$ we have: $$[MBC] = \frac{[ABC]}{2} \implies \frac{BC \cdot BM \cdot \sin \angle MBC}{2} = \frac{BC \cdot h_a}{4} $$
$$\implies \sin \angle MBC = \frac 12 \implies \angle MBC = \frac{\pi}{6} $$
Similarly:
$$[MBA] = \frac{[ABC]}{2} \implies \frac{BA \cdot BM \cdot \sin \angle MBA}{2} = \frac{BA \cdot h_c}{4} $$
$$\implies \sin \angle MBC = \frac 12 \cdot \frac{h_b}{h_a} \le \frac 12 \implies \angle MBC \le \frac{\pi}{6} $$
Therefore $\angle B = \angle BMA + \angle BMC \le \frac{\pi}{3}$
