Fun with an Acute Triangle, its Circumcircle, Altitudes, and Intersections

Just when you thought that it was impossible for me to find more fun in mathematics, I pull this one out. I know that the title is so exciting that you can't wait to read all about it, but be patient and try to take it all in.

In this investigation, I am to construct an acute triangle, ABC, and its circumcircle. Next, the altitudes of the triangles are to be constructed and extended so that they intersect the circumcircle. Let's go on and take care of that.

Of course, you can only see the acute triangle, circumcircle, altitudes, and intersections. In order to construct the circumcircle, I constructed the perpendicular bisectors of the triangle and then constructed the circle using the point of intersection of the perpendicular bisectors as the center of the circle and a vertex on the triangle as a point on the circle.

Now, we are ready to move on to the fun stuff! We want to find:

Click here for a GSP sketch to pull, stretch, shrink, and have fun with to see if this will always work. Make sure that while you are playing with this sketch that you see what happens if we had used a right triangle or an obtuse triangle. You will see why it was important to start with an acute triangle!

I know that you thought we were having fun, but the fun is in the proof! Now, we get to prove that this sum of the ratios is always 4!

Of course, along the way, I have found something else that is interesting. The segments constructed from H to the point on the circle are bisected by the side of the triangle that it passes through. Why you may ask? Let's see if we can figure it out! Yippee!

In looking at triangles BFH and CEH, we have similar triangles by Angle-Angle Similarity. Angle BFH and angle CEH are right angles because the lines constructing the angles are perpendicular -- remember? We constructed those lines as altitudes in the beginning. Next, angle BHF and angle CHE are congruent because they are vertical angles. Therefore, we will actually have three pairs of similar triangles using the same argument:

DBFH ~ DCEH

DAFH ~ DCDH

DAEH ~ DBDH

Now, we have a few more sets of congruent angles that follow from the triangles being similar:

It is very important to realize that angle ABQ and angle ACQ intersect arc AQ, which leads to:

Since by the Reflexive Property and because they are both right angles constructed by the original altitudes, we have congruent triangles by the Angle-Side-Angle Postulate as shown below.

Since we now have congruent triangles, we can use the theorem that states the corresponding parts of congruent triangles are congruent to prove that HE = EQ, HF = FR, and HD = DP. Therefore, the segment from H to the point on the circle is bisected by the side of the triangle that it intersects.

Now, I am going to manipulate some ratios to see what happens -- remember, our goal is to find

First, I am going to look at the first ratio. Using segment addition and substitution, I get:

Using the same arguments as above leads to the other ratios:

This leads to a new ratio of:

From the very beginning of this problem, we know that the sum of the ratios is 4. That means:

I feel that we are getting very close. Now, I want to look at the areas of three smaller triangles that would sum to make the area of the larger triangle ABC.

If we divide both sides of the equation by area of triangle ABC, we get:

This proof is now complete and the sum of our ratios will always be 4!!!!

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