Ratios Created by the Orthocenter
By: Lindsey Harrison
Objective: In this exploration we will look at a given acute triangle ABC and construct the
orthocenter. We will then prove two ratio statements concerning the segment lengths on the
triangle created from the perpendiculars used to form the orthocenter. We will then explore the
obtuse triangle and see if the conjectures hold true.
Given information: Given acute triangle ABC. Construct the Orthocenter H. Let points D, E,
and F be the feet of the perpendiculars from A, B, and C respectfully. GSP FILE
Proof Number One
We are first going to prove the following:
We are going to complete the proof by looking in terms of different representations of total area
of triangle ABC. From our basic formulas, we know that the total area of a triangle is given by
½(base)(height). Thus we can represent area in the following manner
1
1
1
2
2
2
Area(∆ABC)= (𝐵𝐶)(𝐴𝐷) = (𝐴𝐵)(𝐶𝐹) = (𝐴𝐶)(𝐵𝐸).
We can also represent the area of ∆ABC by summing the area of three smaller triangles that
comprise ∆ABC.
Thus we can represent total area as follows:
1
1
1
Area(∆ABC)=2 (𝐴𝐵)(𝐻𝐹) + 2 (𝐵𝐶)(𝐻𝐷) + 2 (𝐴𝐶)(𝐻𝐸).
Observe the following:
1
1
1
𝐴𝑟𝑒𝑎(𝐴𝐵𝐶) 2 (𝐴𝐵)(𝐻𝐹) + 2 (𝐵𝐶)(𝐻𝐷) + 2 (𝐴𝐶)(𝐻𝐸)
=
𝐴𝑟𝑒𝑎(𝐴𝐵𝐶)
𝐴𝑟𝑒𝑎(𝐴𝐵𝐶)
1
1
1
(𝐴𝐵)(𝐻𝐹)
(𝐵𝐶)(𝐻𝐷)
(𝐴𝐶)(𝐻𝐸)
2
2
1=
+
+2
𝐴𝑟𝑒𝑎(𝐴𝐵𝐶) 𝐴𝑟𝑒𝑎(𝐴𝐵𝐶) 𝐴𝑟𝑒𝑎(𝐴𝐵𝐶)
1
1
1
(𝐴𝐵)(𝐻𝐹)
(𝐵𝐶)(𝐻𝐷)
(𝐴𝐶)(𝐻𝐸)
2
2
1=
+
+2
1
1
1
(𝐴𝐵)(𝐶𝐹)
(𝐵𝐶)(𝐴𝐷)
(𝐴𝐶)(𝐵𝐸)
2
2
2
1=
𝐻𝐹 𝐻𝐷 𝐻𝐸
+
+
𝐶𝐹 𝐴𝐷 𝐵𝐸
By the commutative law of addition we have
Proof Number Two
We will still be working with the same acute triangle ABC, but now we are going to prove the
following:
From the previous proof we have the following information we can use:
1=
𝐻𝐹 𝐻𝐷 𝐻𝐸
+
+
𝐶𝐹 𝐴𝐷 𝐵𝐸
We can manipulate this in the following manner since a line segment can be viewed as the sum
of its parts.
1=
𝐶𝐹 − 𝐶𝐻 𝐴𝐷 − 𝐴𝐻 𝐵𝐸 − 𝐵𝐻
+
+
𝐶𝐹
𝐴𝐷
𝐵𝐸
Through algebra we have
1=
𝐶𝐹 𝐶𝐻 𝐴𝐷 𝐴𝐻 𝐵𝐸 𝐵𝐻
−
+
−
+
−
𝐶𝐹 𝐶𝐹 𝐴𝐷 𝐴𝐷 𝐵𝐸 𝐵𝐸
1=1−
𝐶𝐻
𝐴𝐻
𝐵𝐻
+1−
+1−
𝐶𝐹
𝐴𝐷
𝐵𝐸
𝐶𝐻 𝐴𝐻 𝐵𝐻
1=3−(
+
+
)
𝐶𝐹 𝐴𝐷 𝐵𝐸
−2 = − (
2=
𝐶𝐻 𝐴𝐻 𝐵𝐻
+
+
)
𝐶𝐹 𝐴𝐷 𝐵𝐸
𝐶𝐻 𝐴𝐻 𝐵𝐻
+
+
𝐶𝐹 𝐴𝐷 𝐵𝐸
By the commutative law of addition we have
Obtuse Triangle Exploration
When we use our previous GSP file to explore obtuse triangles it appears that the orthocenter
disappears. This however is not the case. Take a look at the following:
GSP File
The orthocenter for an obtuse triangle lies outside of triangle ABC. Not all intersections of the
altitudes and the triangle are there because they do not lie on the line segments of the triangle. It
is always the case that two of the three altitudes either intersect at a vertex (the case for a right
triangle) or they lie outside of the original triangle. We could then no longer talk about the area
of the smaller triangles formed and thus our first proof would not hold. In consequence our
second proof would not hold since it relies on the conclusion of the first.