Assignment # 8
Explorations with Altitudes and Orthocenters
by Michael Ferra
Proposed Investigation
i. Given ▵ABC, construct the orthocenter H. Let points D, E, F be the feet of the perpendiculars from A, B, and C respectfully.
ii. Prove:
iii. Prove:
iv. Given △ABC to be an obtuse triangle, show if these two proofs hold.
i. Given ▵ABC, construct the orthocenter H. Let points D, E, F be the feet of the perpendiculars from A, B, and C respectfully.
-Let's first start out by constructing △ABC such that it is not an obtuse triangle.
-Next let's construct a perpendicular line through vertex A of △ABC to its opposite side BC.
Observe that the perpendicular line intersects side BC. The point at which they intersect is the foot of the perpendicular from A. We will call this point D. The segment AD now defines the altitude of △ABC with a base length defined by segment BC. The area of △ABC can therefore be calculated by A = (1/2)(b)(h), where b represents the base length and h represents the altitude of the triangle in which you wish to find the area. This is the simplest, most common, formula for computing the area of a triangle. Thus for our triangle here we have:
-Similarly, let's construct a perpendicular line through vertex B of △ABC to its opposite side AC.
Observe that the perpendicular line intersects side AC. The point at which they intersect is the foot of the perpendicular from B. We will call this point E. The segment BE now defines the altitude of △ABC with a base length defined by segment AC. Thus:
-We don't want vertex C to feel left out so let's construct a perpendicular line through vertex C of △ABC to its opposite side AB.
Observe that the perpendicular line intersects side AB. The point at which they intersect is the foot of the perpendicular from C. We will call this point F. The segment CF now defines the altitude of △ABC with a base length defined by segment AB. Thus:
-Let's pull all of these illustrations together and see if we notice anything of interest.
Observe that the three perpendicular lines created through the vertices to there respected opposite sides, are concurrent at a point. This point of concurrency is the orthocenter of △ABC, which we will call H.
-Summing up what we've done thus far, we know
If
then
Knowing this could prove to be useful later.
-Let's continue to examine △ABC and the constructions we've made thus far. Is there anything useful we can get from these constructions to help us with our proofs in the next two sections?
Notice that the 3 perpendiculars have subdivided △ABC into six sub-triangles.
For convenience, let's combine the two triangles along segment BC, combine the two triangles along segment AC, and combine the two triangles along segment AB, thus creating 3 sub-triangles of △ABC.
Notice these three triangles have a common vertex at H, the orthocenter. These triangles are △HBC, △HAC, and △HAB. The sum of the areas of these three triangles make up the area of △ABC, thus we can write this as:
Take into account the length of the base and altitude of each of the 3 sub-triangles to give us:
-Now we have a few useful tools to help us in the next few sections.
ii. Prove:
Without loss of generality, I will use the following two area equations to begin this proof:
Since the two equations are equivalent, we can set them equal to one another and go from there.
▩
iii. Prove:
We just proved
We can show by the addition of segments that,
▩
iv. Given △ABC to be an obtuse triangle, show if these two proofs hold.
Let's create △ABC such that it is obtuse and construct the points D, E, F and H in the same manner as outlined in the first section.
