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:
Click here to see my GSP construction with this
calculation.
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:
AH = AD – AH
BH = BE – HE
CH = CF – HF
Substitute
into
And
we get
Recall
from the previous proof:
Substitute
and we get:
3 – 1 = 2
Therefore:
Click here to see my GSP construction with this
calculation.
If you manipulate ABC so that it is an obtuse triangle, you will see that this
manipulation no longer holds. Why
is this?