Similar triangles and orthocenter

Question

$\triangle ABC$ is an acute triangle. $AM$ $(M\in BC)$ and $BN$ $(N\in AC)$ are altitudes. If $CM:AC=1:3$, I should find $NM:AB$ and $CN:BC$.

I was able to show that $\triangle AMC \sim BNC$. Therefore, $\dfrac{AM}{BN}=\dfrac{CM}{CN}=\dfrac{AC}{BC}$. From here we can get $\dfrac{CM}{AC}=\dfrac{CN}{BC}=\dfrac{1}{3}$. What should I do to find $NM:AB$?

Answer 1

Hint: Use angle chasing to show $\triangle CNM\sim \triangle CBA$.

Observe that since the quadrilateral $ABMN$ is cyclic, we have $$\angle MNC=180^\circ-\angle ANM=\beta$$ Similarity implies $$\frac{NM}{AB}=\frac{CM}{AC}=\frac13$$