The orthocenter of a triangle

by Hwa Young Lee

In assignment 4, we investigated the centroid of a triangle and found some properties of the medians. Especially, we found the ratio of the medians divided by the centroid and the area of the small triangles divided by the three medians.

Here, we are going to investigate the orthocenter of a triangle and find some interesting properties of the orthocenter of an acute triangle in terms of the ratio of lengths of segments. The orthocenter of a triangle (usually denoted as H), is the intersection point of the three lines containing the altitudes of the triangle.

An altitude of a triangle is the perpendicular segment joining the vertex to the opposite side.

Let's find out where the orthocenter is for three different kinds of triangles:

a) An acute triangle ABC:  the orthocenter H is inside the triangle.

b) A right triangle ABC: the orthocenter H is on the vertex with the right angle.

c) An obtuse triangle ABC: the orthocenter H is on the lines extended along the altitude, outside of the triangle.

Let's take a closer look at the orthocenter of an acute triangle ABC.

Before we do this, let's construct a circumcircle of triangle ABC and name the intersection of the lines containing the altitudes with the circumcircle points P, Q, and R as in the following diagram:

I'll give you a moment to figure out the following sum of ratios. You can use the GSP construction to explore.

1.

2.

3.

Any guesses?

Now, here are the answers and proofs.

1.

Proof: We use the properties of perpendicular lines and the areas of triangles.

The area of triangle ABC can be expressed in three different ways:

Also, the area of triangle ABC is the sum of the three colored sub triangles.

The area of each colored sub triangles are

, ,.

So,

, by carefully selecting the area of ABC for each term to obtain our desired result. Now we cancel out and obtain:

And our proof is done!

2.

Proof: We know that .

Then

From our result in 1. (above), , thus  and our proof is done!

3.

Why does this hold? Here, I will only prove the first equality . The other two equalities follow the same idea.

If we connect points P and C with a segment, we obtain two triangles (in gray). We are going to show that these two triangles are congruent to prove .

because they are both inscribed angles of the same arc .

Also, so .

Meanwhile, . Hence, .

From , we know that so .

Since .

is the common side. Therefore, , which leaves us with , as desired.

We know that .

Then

, by (*).

From our result in 1. (above), ,

thus  and our proof is done!