Orthocenter Ratios

By: Laura Lowe

Problem: Given triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Prove:

 Prove:

Proof:

We can find the area of ABC three ways:

since AE is the altitude of CB, DB is the altitude of AC, and FC is the altitude of AB.

We can also find the areas of the triangles formed by constructing the perpendiculars.

ABC is divided into 6 smaller triangles by the altitudes BD, AE, and FC.  You can also see that ABC can be divided onto 3 triangles, CHB, AHC, and BHA.  Since BDAC, and H is on segment BD, HDAC as well, so we can find the area of AHC.

Similarly,

We can express ABC = CHB + AHC + BHA.  If we divide both sides of the equation by ABC, we get,

We can substitute the formulas for the areas of ABC, CHB, AHC, and BHA that we found earlier and we get:

Cancel like terms and we get:

If you manipulate ABC so that it is an obtuse triangle, you will see that this manipulation no longer holds.  Why is this?

 Prove:

Proof:

BH = BE – HE

CH = CF – HF

Substitute into

And we get

Recall from the previous proof:

Substitute and we get:

3 – 1 = 2

Therefore: