**Altitude
Ratios**

Suppose I construct an acute triangle ABC. I
then locate the orthocenter H by constructing the altitudes from each vertex.
The orthocenter is the intersection of these altitudes. I then label the feet
of the altitudes D, E, and F. Below is a picture of this construction.

I am interested in the sum of the following
ratios:

(HD)/(AD)
+ (HE)/(BE) + (HF)/(CF)

Using the measure tool in GSP I can calculate
the sum of the ratios of these segments. I find that for any shape of the acute
triangle the sum of the ratios always equals one.

Click here for a
GSP file to try yourself. Why is this so?

Well, let's consider one of the ratios
initially. The segment AD is of course an altitude of triangle ABC while the
segment HD is the altitude of triangle HBC. Since these two triangles share the
same base, the ratio of the areas of HBC to ABC is equal to the ratio of the
segments HD to AD.

It follows then that we can apply this same logic to the ratios (HE)/(BE) and (HF)/(CF) to write:

and

So we have the following picture where the
yellow triangle is HBC, the blue triangle is HAC, and the red triangle is HAB:

So

=

= 1.

Now consider the sum of the ratios:

(AH)/(AD)
+ (BH)/(BE) + (CH)/(CF)

Again using the measure tool in GSP I can
calculate the sum of the ratios of these segments. I find that for any shape of
the acute triangle the sum of the ratios always equals two.

Click here for a
GSP file to try yourself. Let's try and prove why this is so.

Again we'll examine a single ratio and then
apply it to the other two ratios So based on the logic above, I want to try and
relate the ratio (AH)/(AD) to the ratio of some areas of triangles. . I know
that AH is AD-HE. So I can think of AH as being analogous to the area of ABC
minus the area of HBC. Visually it looks like this:

So applying this logic:

(AH)/(AD) + (BH)/(BE) + (CH)/(CF) is equal to

which is equal to

which is equal to

which is equal to

which is equal to 3-1 = 2.

The reader perhaps has observed that none of
the above holds when triangle ABC is obtuse. When this is the case the
orthocenter H is located outside of ABC. We in fact see that the more obtuse
ABC is, the further outside the triangle the orthocenter is located, the
greater the sums of the ratios of segments discussed above.