<s>[INST] 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!
 [/INST] 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
$$\color{red}{AC:CB=3:8}$$
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