Orthocentre of triangle and related ratio $ABC$ is a triangle with $AB = 13$, $BC = 14$ and $CA = 15$. $AD$ and $BE$ are the altitudes from $A$ to $B$ to $BC$ and $AC$ respectively. $H$ is the point of intersection of $AD$ and $BE$. Then the ratio $\frac{HD}{HB}$ is?
I tried by first finding the length of the altitude AD because its possible to find the area. But then I'm unaware of any method of bringing the orthocentre into the picture. How should I do so?


With the help of the image above note that $$\frac{HD}{HB}=\sin(\angle EBC)=\frac{EC}{BC}=\cos(\angle BCA) = \cos\gamma$$
For the Law of Cosines you have $$AB^2=BC^2+AC^2-2BC\cdot AC\cdot\cos\gamma \Rightarrow \\ \cos\gamma = \frac{HD}{HB} = \frac{AB^2-BC^2-AC^2}{-2BC\cdot AC}=\frac{13^2-14^2-15^2}{-2\cdot14\cdot15}=\frac35$$
