i) Triangles $E M B$ and $K A B$ are similar because
$$M B : A B = E B : K B.$$
and the angle at $B$ is shared. Therefore $K \hat{A} B = E \hat{M} B$ and $C A \parallel M H$.

ii) By Thales' theorem we have $C B = 2 H B$, from which we deduce that $C H = H B$ and that $A H$ is the median relative to $C B$. Since $A H$ by hypothesis is also the altitude, the triangle $A B C$ is isosceles.

iii) Since $B D = 2 D K$, the centroid of $A B C$ lies on the line $r$ passing through $D$ and parallel to $A C$ (by Thales' theorem). On the other hand, the centroid also lies on the median $A H$, and therefore the centroid is the point $D$ of intersection of these two lines (note that the two lines are not parallel, since $A C$ is a side and $A H$ is a median of the triangle $A B C$). Therefore $B K$ passes through the centroid and hence is a median. Since $B K$ is also a bisector, $B A = B C$.

iii) $B K$ is the median relative to $C A$. Indeed, suppose by contradiction that the midpoint of $C A$ is $K' \neq K$. Since the intersection point of $B K'$ with $C M$ (which we call $D'$) is the centroid of $A B C$, we would have
$$\frac{K' B}{D' B} = \frac{3}{2} = \frac{K B}{D B}$$
Therefore, since $K \hat{B} K' = D \hat{B} D'$, the triangle $D D' B$ would be similar to $K K' B$ and $C M$ would be parallel to $C A$, which is absurd. We deduce that $K$ and $K'$ are the same point and that $B K$ is a median. Since by hypothesis $B K$ is also a bisector, $A B C$ is isosceles also at $B$ and therefore is equilateral.