Proof involving altitude of a right triangle and circle Let $AD$ be the altitude corresponding to the hypotenuse $BC$ of the right triangle $ABC$. The circle of diameter $AD$ intersects $AB$ and $M$ and $AC$ at $N$. Prove that $\displaystyle\frac{BM}{CN} = \left(\frac{AB}{AC}\right)^3$.
This is what I have so far:
Power of a point(B) = $BM = \displaystyle\frac{BD^2}{AB}$ and Power of a point(C) = $CN = \displaystyle\frac{CD^2}{AC}$. 
Using the Altitude Theorem we in $\triangle ABC$ with altitude $AD$ we obtain: $AD^2 = BD \cdot CD$ and therefore $BD^2 = \displaystyle\frac{AD^4}{CD^2}$ and $CD^2 = \displaystyle\frac{AD^4}{BD^2}$. 
Plugging this into the equations for $BM$ and $CN$ we get:
$BM = \displaystyle\frac{AD^4}{CD^2} \cdot \frac{1}{AB}$ and $CN = \displaystyle\frac{AD^4}{BD^2} \cdot \frac{1}{AC}$. 
I am not sure how to obtain $\displaystyle\frac{AB^3}{AC^3}$. If someone could provide me with a hint as to where to go from here, or if what I have done so far is not the right way to approach the proof please guide me in the right direction.


$$BM=\frac{AD^4}{CD^2}\cdot\frac1{AB}\quad CN=\frac{AD^4}{BD^2}\cdot\frac1{AC}$$
$$\frac{BM}{CN}=\frac{BD^2}{CD^2}\cdot\frac{AC}{AB}$$
Now since
$$\frac{AB}{AC}=\frac{BD}{AD}=\frac{AD}{CD}$$
we have
$$\frac{AB^2}{AC^2}=\frac{BD}{CD}$$
$$\frac{AB^4}{AC^4}=\frac{BD^2}{CD^2}$$
$$\frac{BM}{CN}=\frac{AB^4}{AC^4}\cdot\frac{AC}{AB}=\left(\frac{AB}{AC}\right)^3$$
