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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!

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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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(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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  • $\begingroup$ @tommy xu3. I'm sorry, but I don't get why △XKZ∼△ABZ. Could you elaborate? $\endgroup$ – Albus Dumbledore Jan 1 '17 at 6:55
  • $\begingroup$ $\triangle{XKZ}$ and $\triangle{ABZ}$ should not be similar since $\angle{XKZ}=90^\circ$ and $\angle{ABZ}$ is obtuse. $\endgroup$ – mathlove Jan 1 '17 at 7:52
  • $\begingroup$ You are right. I misunderstood the problem statement as $YB\perp AB$ Sorry. $\endgroup$ – User Jan 1 '17 at 8:00

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