Find the height drawn from the foot of interior bisector in an obtuse triangle The problem is as follows:

A $\triangle ABC$ is obtuse on $\angle B$. The interior bisector $BM$ is
traced from point $B$ and as well the altitudes $AN$ and $CQ$
respectively. Assuming that $AN=8\,cm$ and $CQ=12\,cm$. Find the
length of the altitude traced from $M$ in triangle $\triangle{BMC}$.

The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&8\,cm\\
2.&6\,cm\\
3.&5.4\,cm\\
4.&4.8\,cm\\
\end{array}$
The figure from below is the interpretation which I could conclude from reading the word problem, however that's it. I don't know what else can be established from there?.

I think that it requires, similarity or congruence but I cannot say for sure exactly where should it be applied. Please include a drawing or a diagram in the answer.
So far can this be solved relying only in Euclidean postulates?.
 A: First of all, we should all agree that $\overline{MH}=\overline{MG}$ since $\overline{MB}$ is an angle bisector of $\angle ABC$.
Now since $\overline{MH}\parallel\overline{AN}$, we know $\triangle MHC$ and $\triangle ANC$ are similar to each other by A.A.A. Hence we have $\frac{\overline{MH}}{\overline{AN}}=\frac{\overline{MC}}{\overline{AC}}$.
Similarly, since $\overline{MG}\parallel\overline{CQ}$, we know $\triangle AMG$ and $\triangle AQC$ are similar to each other by A.A.A. Hence we have $\frac{\overline{MG}}{\overline{CQ}}=\frac{\overline{MA}}{\overline{AC}}$.
Therefore
$\frac{\overline{MH}}{\overline{AN}}+\frac{\overline{MG}}{\overline{CQ}}=\frac{\overline{MC}}{\overline{AC}}+\frac{\overline{MA}}{\overline{AC}}=1\\
\Longrightarrow \frac{\overline{MH}}{8}+\frac{\overline{MG}}{12}=1\Longrightarrow \frac{1}8+\frac{1}{12}=\frac{1}{\overline{MH}}\Longrightarrow \overline{MH}=\color{red}{4.8\text{ (cm)}}$

A: 
$\triangle ANB \sim \triangle CQB$ by $AA$ similarity with ratio $$r=\frac{8}{12}=\frac{2}{3}$$
It follows
$$  \dfrac{AB}{CB} = \dfrac{2}{3} = \dfrac{AM}{CM}$$
as angle bisector divides the base into ratio of adjacent sides.
And as $MH || AN$, $$  \dfrac{MH}{AN} = \dfrac{CM}{AC} = \dfrac{CM}{AM + CM} =\dfrac{3}{2 + 3}=\dfrac{3}{5}$$
$$\therefore \boxed{MH =\dfrac{3}{5}AN =4.8}$$
A: First, let $x$ be the length of the altitude traced from $M$ in triangle $\triangle BMC$, and name $R$ the point where this altitude intersects $\overline{BC}$. If we look at $\triangle BMA$, let $y$ be the length of the altitude traced from $M$ to $\overline{AB}$, which intersects this segment at the point $S$. We can see that $x=y$, because $\triangle MRB \cong \triangle MSB$ (using AAS rule).
Looking at $\triangle ANC$ and applying Thales' theorem, we can state that $$\frac{\overline{MR}}{\overline{AN}} = \frac{\overline{MC}}{\overline{AC}} \iff \frac{x}{8} = \frac{\overline{MC}}{\overline{AC}}.$$
Analogously for $\triangle AQC$, we get $$\frac{y}{12} = \frac{\overline{AM}}{\overline{AC}}.$$
Summing up both equations, we get $$\frac{x}{8} + \frac{y}{12} = \frac{\overline{AM} + \overline{MC}}{\overline{AC}}$$ $$\iff \frac{x}{8} + \frac{x}{12} = \frac{\overline{AC}}{\overline{AC}}$$ $$\iff \frac{5}{24}x = 1$$ $$\iff x = \frac{24}{5} = 4.8 \, \textrm{cm}.$$
