Distance between orthocenters Given a triangle $\Delta ABC$ with angle $\angle ACB=90^\circ$. The bisector of $\angle ACB$ intersects $AB$ at $L$. If $AC=4 \operatorname {cm}$. and $BC=5 \operatorname {cm}$, find the distance between the orthocenter $H_1$ of $ALC$ and the orthocenter $H_2$ of $BLC$.

A fun approach, using a bit of geometry:
Let $a = AC$ and $b = BC$.  Assume that $a < b$, as in your example.  Let $D$ be the intersection of the altitude from $L$ to $AC$.  Let $E$ be the intersection of the altitude from $L$ to $BC$.  (Sorry for the lack of a diagram, but I don't have a scanner available, and I don't want to teX it out.)  Since $\angle DCL = 45^{\circ}$, it's easy to see that $CD = DL = EL = CE$.  Let $x$ be the common length.  By similar triangles ($\triangle ADL \sim \triangle LEB$), we have
$$ \frac{a-x}{x} = \frac{x}{b-x}  \;\Rightarrow\; x = \frac{ab}{a+b}.$$
Furthermore, $\angle ADH_1 = \angle BEH_2 = 45^{\circ}$ (which is easy enough to verify by labeling angles).  Therefore $AD = DH_1$ and $BE = EH_2$.  Consider $\triangle H_1LH_2$.  Note, $H_1$ is in the interior of $\triangle ABC$, $H_2$ exterior, and $\angle H_1LH_2 = 90^{\circ}$.  Thus, the required length is found by the Pythagorean Theorem:
$$\begin{align*}
(H_1H_2)^2 &= (LH_1)^2 + (LH_2)^2 \\
&= (LD - DH_1)^2 + (EH_2 - LE)^2 \\
&= \left( \frac{ab}{a+b} - \left[a - \frac{ab}{a+b}\right]\right)^2
+ \left( \left[b - \frac{ab}{a+b}\right] - \frac{ab}{a+b}\right)^2\\
&= \textrm{a few steps of algebra...}\\
&= \left(\frac{b-a}{a+b}\right)\sqrt{a^2 + b^2}.
\end{align*}
$$
I like the formula, as it involves a relationship between the sum and difference of the legs, and the hypotenuse of the triangle.
