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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 $BLC$.

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2 Answers 2

If one draws a few of the obvious lines, some nice similar triangles appear. But we resolutely close our eyes to all that, and show how a crude sledgehammer approach works.

The triangle is said to have the two legs $4$ and $5$. Put $C$ at the origin, with $A$ at $(5,0)$ and $B$ at $(0,4)$. So the line $AB$ has equation $4x+5y=20$.

We find the coordinates of the orthocentres of $\triangle ALC$ and $\triangle BLC$. Each of these triangles has the origin as a vertex. Let us find the equation of the line through the origin which is perpendicular to $AB$. (This line is an altitude to both of our small triangles.)

The slope of $AB$ is $-4/5$, so our line has slope $5/4$, and therefore equation $y=5x/4$.

Now we find the equation of the altitude of $\triangle ALC$ that passes through $L$. This line is vertical, so all we need is the $x$-coordinate of $L$. The line $CL$ has equation $y=x$. Substituting in $4x+5y=20$, we get $x=20/9$.

So the orthocenter of $\triangle ALC$ is where the line $x=20/9$ meets the line $y=5x/4$. This intersection point $H_1$ is easy to find.

Go through the same procedure with $\triangle BLC$, and find the coordinates of $H_2$. Then compute the distance between $H_1$ and $H_2$.

Remark: The method of coordinates usually attributed to Fermat and Descartes is very powerful. When we marry it to the computer algebra packages now available, we obtain an essentially mechanical procedure for solving geometric problems of the traditional kind.

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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.

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