<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] (Wrong!!! This is the case for $YB\perp AB.$)
Following your attempts, we get $\frac{11}{3}\times\frac{AC}{CB}\times\frac{YB}{7} = 1.$ Hence it remains to find the length of $YB.$ Let $K$ be the point on $YZ$ such that $XK\perp YZ.$ Then we can use the area of the triangle to find $XK$:
$$\sqrt{10(10-9)(10-7)(10-4)}=\frac{7\times XK}{2}.$$
Therefore $XK=\frac{12\sqrt{5}}{7}.$
Then since $\triangle XKZ\sim\triangle ABZ,$ we can get
$$AB = XK \times\frac{63/11}{9}=\frac{12\sqrt{5}}{11}$$
and obtain $BZ = \sqrt{(\frac{63}{11})^2-(\frac{12\sqrt{5}}{11})^2}=\frac{57}{11}.$
Then our $YB=7- \frac{57}{11} = \frac{20}{11},$ so the answer should be
$$\frac{AC}{CB} = \frac{3}{11}\times\frac{7}{YB} = \frac{21}{20}.$$
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