**Altitudes of a Triangle**

**By**

**Janet M. Shiver**

EMAT 6680

**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.

This time
we will look at a right triangle. It appears that the three altitudes
are now intersecting in a single point at one of the vertices.

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.

Proof:

ConsiderD ABC with altitudes
By definition of an altitude of a triangle,
is perpendicular to
,
is perpendicular
to
and
is perpendicular to
.

**Construct
****||****
and passing through point A.**

Similarly, construct
||
and passing through point C and
||
and passing through point B.

Since
||
and
is a transversal through these segments, angle AQB is congruent
to angle DAQ by the congruence of alternate interior angles.
In addition, by the definition of perpendicular angle AQB is a right angle.
Thus angle DAQ must also
be a right angle. Therefore,
is perpendicular to
by definition of perpendicular lines.

Using a similar argument, we can show
that
is perpendicular to
and
is perpendicular to
.

Next we will show that each of the
altitudes is a segment bisector. By construction
||
and
||
. By definition of a parallelogram, ABCD forms a parallelogram and thus
AD=BC.

By construction
||
and
||
.By definition of a parallelogram, ACBE forms a parallelogram and thus EA=BC.

Using the transitive property
of equality, EA = AD. By definition of midpoint, A is the midpoint
of
and separates the segment
into two congruent parts
and
. Thus,
is the perpendicular bisector of
.

We can use a similar argument to show
that
is the perpendicular bisector of
and that
is the perpendicular bisector of
. Since we have shown that each segment
,
and
is a perpendicular bisector then all three segments must be concurrent by
the theorem for the concurrence of the perpendicular bisectors of a triangle.

Therefore, since the concurrent perpendicular
bisectors of the sides of triangle DEF are also the altitudes of triangle
ABC, these altitudes must be concurrent.

Return to Janet Shiver's Home Page