The point H is the orthocenter of the triangle ABC and the point C is the centroid of the triangle ABH. In that case the smallest angle of the triangle ABC is: (60), (30), (45), ($\angle ACB$)?
This is actually quite tough. I got that the orthocenter is the meeting of all of the altitudes, but I still can't figure this out.
And $c$ is the median of $\triangle ABH$. I don't see anyway to proceed.
Hint:
$ABH$ is an equilateral triangle, $C$ is its center, so $H$ is the orthocenter of $ABC$.
So the smallest angle of $ABC$ is $30°$.
Using $$\vec{AH}=\frac{\tan B\vec{AB}+\tan C\vec{AC}}{\tan A+\tan B+\tan C},$$ (see the answer for this question) we have $$\vec{AC}=\frac{\vec{AB}+\vec{AH}}{3}$$ $$\Rightarrow \vec{AC}=\left(\frac{1}{3}+\frac{\tan B}{3(\tan A+\tan B+\tan C)}\right)\vec{AB}+\frac{\tan AC}{3(\tan A+\tan B+\tan C)}\vec{AC}.$$
Hence, we have $$0=\frac{1}{3}+\frac{\tan B}{3(\tan A+\tan B+\tan C)}\quad\text{and}\quad 1=\frac{\tan C}{3(\tan A+\tan B+\tan C)},$$ i.e. $$0=\tan A+2\tan B+\tan C\quad\text{and}\quad 3\tan A+3\tan B+2\tan C=0,$$ i.e. $$\tan A=\tan B=a,\tan C=-3a\ \text{for some $a$}$$
Since it is known that $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$for $A+B+C=\pi$ (see this question), we have $$a+a-3a=-3a^3.$$ Hence, $a=\frac{1}{\sqrt 3}$ gives $A=B=30^\circ, C=120^\circ.$
See triangleabh is equilateral hence abc has orthocentre h hence smallest angle is 30 degrees
