Orthocenter and Altitude Ratios
By Courtney Cody
In this assignment, I will investigate the altitude ratios created by a triangleŐs orthocenter. Specifically, given triangle ABC, I will construct the orthocenter H, let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully, and prove the following equations:
Consider the following triangle ABC:
Now, letŐs construct the altitude for all three vertices and let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. We will mark the intersection of these altitudes as point H. Recall that a triangleŐs orthocenter is the intersection of these three altitudes. Hence, we have the following triangle with orthocenter H:
In order to help with our proof, we will divide triangle ABC into three smaller triangles, namely ACH, ABH, and ACH. These are pictured below in orange, purple, and green, respectively.
Since we know these three smaller triangles make up the larger triangle ABC, we have the following equation:
Dividing the entire equation by , we have:
Since the altitudes meet the sides of triangle ABC at right angles, then we can determine the areas of all triangles in the diagram. Also, we can express the area of triangle ABC in three different ways since there are three different altitudes and corresponding sides, which we will carefully use since we are trying to form a particular sum of ratios. Consider the following:
which reduces to
Thus, we have established the first equation.
In order to prove the second equation, we will begin with the first equation and make necessary modifications. To begin, we will use the drawing to express the segments represented by the numerators of each proportion in the first equation as the difference of two other segments as follows:
Substituting these new representations into the first ratio, we have the following equation:
Thus, we have established the second equation.
Recall that our proofs of two equations were based on a given acute triangle. What if ABC is an obtuse triangle? Will our equations still hold? LetŐs think about this! If we drag point C such that angle CAB is obtuse (greater than 90 degrees), then we notice that the orthocenter H moves outside of triangle ABC. See the diagram below:
If the triangle is obtuse, then it is obvious that points E and F disappear from the diagram since their corresponding altitudes do not intersect the sides of triangle ABC. Hence, segments BE and CF are no longer relevant, which means that the two equations proved in this exploration do not apply to this diagram. Thus, the two equations we proved do not hold when triangle ABC is obtuse.