geometry
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.
Construct the equilateral triangle $AHB$. Given that $AC = BC, AH = BH$ and the shared $CH$, the triangles $AHC$ and $BHC$ are congruent. Then, $\angle BCH = \dfrac12\angle ACB = 20^\circ$.
Since $AH = BH$ and $\angle BAM = \angle HAM = 30^\circ$, the triangles $BAM$ and $HAM$ are congruent, which yields $\angle HBM = \angle BHM = \angle HBC = 10^\circ$ and $HM || CB$.
Then, the triangles $CHB$ and $BHC$ have the same altitudes $h$ with respect to the base $BC$. Since $\angle BCH = \angle CBM = 20^\circ$, we have $CH = BM = h\cot 20^\circ$.
As a result, the triangles $CHB$ and $BMC$ are congruent, which leads to,
$$\angle BMC = \angle CHB = 180^\circ - \angle CBH - \angle BCH = 180^\circ - 10^\circ - 20^\circ = 150^\circ$$
As we discussed in the comments, $H$ is the orthocenter and then simple angle chasing gives $\angle HAC = 40^0$.
Alternatively, by Trigonometric form of Ceva's theorem, we have
$ \sin (\angle HBC) \ \sin (\angle HCA) \ \sin (\angle HAB) = \sin (\angle ABH) \ \sin (\angle HCB) \ \sin (\angle HAC)$
$\sin 40^0 \sin (60^0-x) \sin 30^0 = \sin 20^0 \sin 30^0 \sin x \ $ ($\angle HAC = x)$
$\sin 40^0 \sin (60^0-x) = \sin 20^0 \sin x$
$\cos (20^0 - x) - \cos(100^0 - x) = \cos (20^0-x) - \cos (20^0+x)$
$\implies x = 40^0$
Best Answer
Hint:
$ABH$ is an equilateral triangle, $C$ is its center, so $H$ is the orthocenter of $ABC$.
So the smallest angle of $ABC$ is $30°$.