Do they meet?
So often we know the statements of theorems pertaining
to the geometry of triangles, but we have never explored why they
are true by proving them. In my experience, I have often take
statements to be true without really knowing why they are. Over
the years, my understanding of the benefits of the proofs of mathematical
theorems has expanded enormously, which has encouraged me to approach
these proofs that are seen here.
What about those Angle Bisectors?
We have all heard that the three angle bisectors ofthe
internal angles of a triangle meet at a point called the incenter.
How do we know that these three lines intersect at the same point?
Here we will prove that the three angle bisectors of
the internal angles of a triangle are concurrent.
Consider the arbitrary triangle ABC, in which the dotted
lines are bisectors of the angles B and C and P is the intersection
of these angle bisectors.
Since z bisects <ACB, the <ACP is congruent to
<BCP by the definition of an angle bisector.
Denote<BCP = c1 and <ACP = c2.
Drop the perpendicular segments from P to all three
sides of triangle ABC. These are the green segments.
Since triangles PFC and PEC are right triangles,
Since c1=c2 as said before, then we see that u=v.
We can make the same argument with right triangles
GPB and EPB to show that w=v. Therefore, u=v=w. This means the
perpendicualr segments from P to the three sides of our triangle
are all congruent. At the end of our proof, we will show what
special circle the three points G, F, and E lie on. Can you make
a guess now?
Now we will draw the segment from A to P. We must show
that this segment bisects angle BAC to show that the angle bisectors
of the three internal angles of a triangle are concurrent.
Let <BAP=b1 and <CAP=b2. We have already shown
that u=w which makes one pair of sides of the triangle congruent,
and we see that they both share the side y. Since triangles APG
and APF are right triangles and we have shown two sides to be
congruent, then GA must equal FA. This would mean that these two
triangles are congruent and therefore all of their angles are
congruent making b1=b2.
This means that AP is the angle bisector of the vertex
A and all three angle bisectors are concurrent!
P is called the incenter of the triangle ABC. This
point is the center of the incircle of which G, F, and E are the
points where the incircle is tangent to the triangle.
Click here to play
with a dynamic GSP file of the illustration of this proof.
A Deeper Look at the Medians
We have also heard that the intersection of the three
medians of a triangle is called the centroid. How do we know that
these three medians intersect at the same exact point?
Here we will prove that the three medians of a triangle
that the point of concurrence, called the centroid,
is two-thirds the distance from each vertex to the opposite side.
For this proof we will place and arbitrary triangle
into the coordinate system and use our algebra skills to prove
each part of the proof.
Let AX and CY be medians of our triangle. (This means
X is the midpoint of BC and Y is the midpoint of AB.) Also let
their intersection be T.
We must prove that the line BT intersectsthe segment
AC at the midpoint of AC, which we will call Z.
Let's look at some algebra to find the equations of
the line passing through A and X and the line passing through
Y and C so that we can calculate their intersection.
We must first find the coordinates of X and Y.
It is clear that X=(c/2, 0). By similar triangles,
Y=(a/2, b/2). These points will help us calculate our lines.
Now since T is the intersection of these two lines,
Now that we have the coordinates of T, we can calculate
the equation of the line passing through B and T and the line
passing through A and C in order to help us find the coordinates
of point Z. If we can prove that Z is the midpoint of AC, then
we will be able to conclude that all three medians pass through
one point and that point is T.
Now we can set these two equations equal to each other
and solve for Z.
Since b/2 is the midpoint of AC (shown by similar triangles),
line BT intersects segment AC at the midpoint of AC, which is
AX and CY are the medians of triangle ABC which intersect at T
and BT intersects AC at it's midpoint, then all three medians
of triangle ABC pass through the common point T, which is called
Both of these proofs have shown us
that given an arbitrary triangle, the three angle bisectors meet
at the incenter and the three medians meet at the centroid.