Right Triangle Problem with Angle Bisector Theorem Right triangle $ABC$ has $AC = 8$ and $CB = 6$. $M$ is the midpoint of $AB$. Pick point $N$ on line $CM$ with $M$ between $C$ and $N$ such that $∠CAB = ∠BAN$. Compute $MN$. Express your answer as a common fraction.

I figured out that $CM$, $BM$, and $AM$ were $5$, but I can't figure out how to continue.
 A: We know $CM = 5$.  
Also let $\angle CAM = a$. Then we have 


*

*$\angle MAN=a$,

*$\angle AMN = 2a$ and 

*$\angle MNA = 180-3a$


Now use the sine rule in $\triangle AMN $ and $\triangle AMC$
$$\dfrac{\sin(180-3a)}{AM} = \dfrac{\sin(a)}{MN}$$
and
$$\dfrac{\sin(a)}{CM} = \dfrac{\sin(180-2a)}{AC}$$
A: I think it is better to leave the trigonometry out of the solution. 
Since $\angle BCA = 90^{\circ}$ one can conclude that $AM = BM = CM$. By Pythagoras one gets $AB = 10$ so  $AM = BM = CM = 5$. Triangle $ACM$ is isosceles and $$\angle \, ACN = \angle \, ACM = \angle\, CAM = \angle \, CAB =  \angle \, BAN = \angle \, MAN$$ Let's look at triangles $ACN$ and $AMN$. We have concluded that $\angle\, ACN = \angle \, MAN$. Since $\angle \, ANM = \angle \, CNA$ as a common angle at vertex $N$, triangles $ACN$ and $AMN$ are similar and therefore
$$\frac{AM}{CA} = \frac{MN}{AN} = \frac{AN}{CN}$$ However, since $AM = 5, \,\,$ $AC=8$ and $CN = CM + MN = 5 + MN$, we get the relations $$\frac{5}{8} = \frac{MN}{AN} = \frac{AN}{5+MN}$$ Thus, one can express $AN$ in two different ways 
$$AN = \frac{8}{5}\, MN \,\,\, \text{ and } \,\,\, AN = \frac{5}{8}\, (MN + 5)$$ Equate the two, obtaining the equation
$$\frac{8}{5}\, MN =  \frac{5}{8} (MN + 5)$$ Solve the equation for $MN$ in order to obtain $$MN = \frac{125}{39}= 3 + \frac{8}{39}$$
