# 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. • Just a hint, thought it might help: Join $MD$ and $ND$. We get a rectangle, Now you may use similarity? – samjoe May 11 '17 at 13:23

$$BM=\frac{AD^4}{CD^2}\cdot\frac1{AB}\quad CN=\frac{AD^4}{BD^2}\cdot\frac1{AC}$$ $$\frac{BM}{CN}=\frac{BD^2}{CD^2}\cdot\frac{AC}{AB}$$ Now since $$\frac{AB}{AC}=\frac{BD}{AD}=\frac{AD}{CD}$$ we have $$\frac{AB^2}{AC^2}=\frac{BD}{CD}$$ $$\frac{AB^4}{AC^4}=\frac{BD^2}{CD^2}$$ $$\frac{BM}{CN}=\frac{AB^4}{AC^4}\cdot\frac{AC}{AB}=\left(\frac{AB}{AC}\right)^3$$
• Why does $\displaystyle\left(\frac{AB}{AC}\right)^2 = \frac{BD}{CD}$ and not $\displaystyle\left(\frac{BD}{CD}\right)^2$ – rover2 May 11 '17 at 14:10
• @rover2 Both $\frac{BD}{AD}$ (1) and $\frac{AD}{CD}$ (2) are equivalent to $\frac{AB}{AC}$ by similar triangles. If I multiply (1) by (2) I get the equality you inquired about. – Parcly Taxel May 11 '17 at 14:12