Quadratic and Cubic
by Jeff Hall
Prove that the lines of the three altitudes
of a triangle are concurrent
First, let's draw a picture of a triangle.
I included the three altitude lines.
The point where the three altitude
lines cross is called the Orthocenter.
It is also a Concurrent Point because
all three lines cross at this single point.
This illustrates that the three altitude
lines are concurrent for this triangle alone.
Now let's prove it for all triangles.
Before we start our proof, we need to devise a strategy.
We know that perpindicular bisectors
are concurrent at the centroid,
so if we could show that the orthocenter
of one triangle is equal to the
centroid of a bigger triangle, we would
prove that the orthocenter is
a point of concurrency for all triangles.
First, using parallel lines, create
a larger triangle around the original.
I labeled the new vertices as A', B',
Notice that A, B, and C are now the
midpoints of their respective segments.
The centroid, or the point of concurrency
for perpindicular bisectors, is found
by connecting each vertex to their
Here is the graph with the centroid
Unfortunately, the centroid and the
orthocenter do not match in the current shape.
But is it possible to redraw the triangle
so that the two points are equal?
Let's try it. Let's move point B to
We now have the centroid and the orthocenter
being the same point.
Since the centroid is proven to be
a point of concurrency for all triangles,
we have now proven that the orthocenter
is a point of concurrency
for all triangles as well.