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.

  • $\begingroup$ Just a hint, thought it might help: Join $MD$ and $ND$. We get a rectangle, Now you may use similarity? $\endgroup$ – 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 $$\frac{AB^2}{AC^2}=\frac{BD}{CD}$$ $$\frac{AB^4}{AC^4}=\frac{BD^2}{CD^2}$$ $$\frac{BM}{CN}=\frac{AB^4}{AC^4}\cdot\frac{AC}{AB}=\left(\frac{AB}{AC}\right)^3$$

  • $\begingroup$ Why does $\displaystyle\left(\frac{AB}{AC}\right)^2 = \frac{BD}{CD}$ and not $\displaystyle\left(\frac{BD}{CD}\right)^2$ $\endgroup$ – rover2 May 11 '17 at 14:10
  • $\begingroup$ @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. $\endgroup$ – Parcly Taxel May 11 '17 at 14:12
  • $\begingroup$ Got it ! Thank you for helping me finish :) $\endgroup$ – rover2 May 11 '17 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.