Find $AC: CB$ in $\triangle XYZ$ Problem:
In $\triangle XYZ$, $XY = 4$, $YZ = 7$, and $XZ = 9$. Let $M$ be the midpoint of $\overline{XZ}$, and let $A$ be the point on $\overline{XZ}$ such that $\overline{YA}$ bisects angle $XYZ$. Let $B$ be the point on $\overline{YZ}$ such that $\overline{YA} \perp \overline{AB}$. Let $\overline{AB}$ meet $\overline{YM}$ at $C$. Find $AC: CB$.
Attempt:
We know that $XM$ = $MZ$ = $\dfrac{9}{2}$, and by the Angle-Bisector theorem, 
$$\dfrac{4}{x} = \dfrac{7}{9-x}$$
$$\implies 36-4x = 7x$$
$$\implies x= \dfrac{36}{11}=XA$$
Therefore, $AZ$ = $\dfrac{63}{11}$, and $AM$ = $\dfrac{9}{2} - \dfrac{36}{11}= \dfrac{27}{22}.$
 Also, by the Menelaus theorem,
$$\dfrac{ZM}{MA} \times \dfrac{AC}{CB} \times \dfrac{YB}{YZ} = 1$$
$$\implies \dfrac{11}{3} \times \dfrac{AC}{CB} \times \dfrac{YB}{7} = 1$$
From here, I got stuck. I'm wondering if one could use mass points, since the problem wants to find the ratio of lengths, and not specific side lengths. Any help is appreciated!

You have already got
$$\dfrac{11}{3} \times \dfrac{AC}{CB} \times \dfrac{YB}{7} = 1\tag1$$
Also, we have
$$YB=\frac{AY}{\cos \frac Y2}\tag2$$
So, we want to find $\cos\frac Y2$ and $AY$.
By the law of cosines, 
$$9^2=4^2+7^2-2\cdot 4\cdot 7\cos Y\implies \cos Y=-\frac 27$$
from which we have
$$\cos\frac Y2=\sqrt{\frac{1-\frac 27}{2}}=\sqrt{\frac{5}{14}}$$
Also, the length of $AY$ is given by $$AY=\sqrt{YX\cdot YZ\left(1-\frac{YZ^2}{(YX+YZ)^2}\right)}=\frac{4\sqrt{70}}{11}$$
(see here or here for this formula)
So, from $(1)$ and $(2)$, we get
$$\color{red}{AC:CB=3:8}$$
