Altitudes of a Triangle
Investigation: Construct the three altitudes of a triangle and prove that they are concurrent.
We will begin our investigation by looking at several triangles and their altitudes.
Here we see an acute triangle. It appears that all three of the altitudes intersect in a single point.
Here we see an obtuse triangle. It appears that the three altitudes of the triangle once again intersect in a single point, however this time they intersect outside the triangle.
Necessary definitions and theorems
Definition of concurrent:
A number of lines are concurrent if they have exactly one point in common.
Definition of altitude of a triangle:
A line segment drawn from a vertex of the triangle perpendicularly to the side opposite the vertex.
Definition of orthocenter:
The point of concurrence for the three altitudes of a triangle.
Theorem of perpendicular bisectors:
The three perpendicular bisectors of the sides of a triangle are concurrent.
Outline of proof:
We will prove that the three altitudes of a triangle are concurrent by constructing three auxiliary lines to form a new triangle and showing that the altitudes of the of the original triangle are perpendicular bisectors of the sides of the newly formed triangle. We will then use the theorem for perpendicular bisectors to prove concurrence.
ConsiderD ABC with altitudes
Using a similar argument, we can show
Next we will show that each of the
altitudes is a segment bisector. By construction
Using the transitive property
of equality, EA = AD. By definition of midpoint, A is the midpoint
We can use a similar argument to show
Therefore, since the concurrent perpendicular bisectors of the sides of triangle DEF are also the altitudes of triangle ABC, these altitudes must be concurrent.