In the current high school curriculum students are taught that a triangle has four points of concurrency; the incenter, orthocenter, centriod, and circumcenter. The students are drilled about the properties of each of these points and the lines that create. Consequently we give students the false assumption that their is no other points of concurrency except for these four points. How can we demonstrate the exsistance of more points of concurrency in a triangle? To do this we begin by extending a concept just taught, in this case the point of concurrecny known as the incenter; so given triangle ABC we find the incenter and knowing the properties of the incenter we can construct a circle that intersects each side of the triangle once.

Next we will construct line segments AF, CE, and BD. We should ask the students a series of questions at this piont to see what they can tell us about the intersection of these lines. It is vital that we have the students understand that this is a point of concurrency but not the incenter, circumcenter, orthocenter, or centriod. To demonstrate this if time permits have the students find the circumcenter, orthocenter, and the centriod to see that this point H is not in fact one of those points.

Are students just going to take our word and
believe us that this is a point of concurrency? We hope not, since in math 1
part of our curriculum includes the writing of proofs. So we will include these proofs in our
lesson. How can we prove that this point H is in fact a point of concurrency?
At this time we need to inform students about a very powerful theorem proven by Giovanni Ceva appropriately named
Ceva's Theorem. This theorem states: "lines from the vertices of a
triangle to the opposite sides are concurrent precisely when the product of the
ratio of the sides are divided is 1 (MacTutor)."

The line from the vertex of a triangle to any
point on the opposite side is called a Cevian. So given a triangle ABC, the point M is a point of
concurrency if and only if. Using CevaÕs
theorem we can now prove that the point H in our triangle is in fact a point of
concurrency. The proof follows:

**Proof that the Gergonne
Point is a Point of Concurrency**

Since G is given as the incenter of the circle if
we construct a line segment BG it must be the angle bisector of ∠ABC thus ∠GBF ≅ ∠GBE. Because of the reflexive property we have that BG ≅ BG. Since GF and GE are radii of a circle and both E and F are tangent points then we have that GF⊥BC and EG⊥BA. Thus we have two right triangles △EGB and △FGB. By the Hypotenuse Angle Congruency theorem we have that △EGF ≅ △FGB.

Since G is
given as the incenter of the circle if we construct a segment CG it must be the
angle bisector of ∠BCA thus
∠GCD ≅ ∠GCF.
Because of the reflexive property we have that GC ≅
GC. Since GF and GD are radii of a
circle and both D and F are tangent points then we have that GD⊥AC and
GF⊥BC. Thus we have two right triangles △FGC and △DGC. By the Hypotenuse Angle Congruency
theorem we have that △FGC ≅ △DGC.

Since G is
given as the incenter of the circle if we construct a segment AG it must be the
angle bisector of ∠BAD thus
∠GAD ≅ ∠GAE.
Because of the reflexive property we have that AG≅AG. Since GD and GE are radii of a circle
and both D and F are tangent points we have that GD⊥AC and GE⊥AB. Thus we have two right triangles △GDA and △GEA. By the Hypotenuse Angle Congruency
theorem we have that △GDA ≅ △GEA.

In order for
the point H to be concurrent that would mean that
. From what we have shown above we
know that AD≅AE, DC ≅ FC, and
BF ≅ BE.
So we can substitute these segments and have thatwhich is all equal to 1. Thus by Ceva's Theorem H is a point of
concurrency since it is the point where the cevians intersect. ∴

Throughout
the proof we want the students to understand how we are using the properties of
the incenter and how using the triangle congruency theorems as a piece of the
proving process instead of it being the end of it. Next we want students to question what we did; we want to
make sure they understand the logic behind the proof and why the proof
works. Our goal in the classroom
is to have students question our process and why things work. We hope to have trained our students
well enough for them to ask us how do we know that CevaÕs Theorem is true; what
is proof for that?

What our
goal is with this proof for CevaÕs theorem is for students to see how the area
formula for a triangle can be used in a proof. In addition, we want students to notice the use of the
triangles within a triangle as many more interesting investigations can be done
in this way. Finally, this proof
will help the students by given them a review of proportions and ratios.

**Proof of CevaÕs Theorem**

**Given: Triangle ABC
with Cevians AD, BF, and CE with M being the point of concurrency**

**Prove: **

We begin by
looking at Cevian AD; and looking at triangle ADC and triangle ADB. We see that both triangles have the
same height (AK), and that their bases share the proportion that we want to
create. Since both triangles have
the same height finding the proportions of these two areas might help:

We also come
to notice that triangle BMD and triangle CMD share the same height ML and that
their bases share the proportion that we want to create.

If we find
the difference of these two areas we will have the ratio of triangles AMB and
AMC. This leads us to the idea
that if we repeat this process around the triangle we will find different
ratios to represent the same areas.
Thus when we multiply these ratios it will equal one.

We now
repeat this process with side BA.

Now we
repeat this process with side BA:

Now let us
summarize what we have found:

Our goal was
to find that From
what we know now we find that:

Thus if the
point M is a point of concurrency then .

Using this
GSP file we have an independent point M to see that this is always true no
matter where point M is inside of the triangle.

CEVA EXPLORE
FILE

The second
page of the document includes a further investigation to this theorem. In the GSP file we have the option of
placing the point M outside of the triangle and we find that the ratio from
CevaÕs theorem still holds. So we
must now demonstrate that CevaÕs Theorem must also be proven for this case,
showing students that some proofs require us to look at multiple cases or
scenarios.

**Continuation of CevaÕs
Theorem**

**Given: Triangle ABC
with Cevians AD, BF, and CE with M being the point of concurrency on the
outside of the triangle**

**Prove: **

Just like
before with the point in the interior of the triangle we want to start by
finding the ratios of the areas and finding the pieces that overlap one
another.

We repeat
this process for the next side of the triangle:

Finally we
repeat this process for the last side of the triangle:

Again let us
review what we have found:

Our goal was
to find that From
what we know now we find that:

Thus even if
point M is on the outside of the triangle .

Giving the
students CevaÕs Theorem again:

ÒIf the
point M is a point of concurrency then .
Ò

Returning to
our standard on the logical statements we remember that given any statement we
can also have the converse, inverse, and contrapositive of the conditional
statement. Students have had
experience with the converse when working with the Pythagorean theorem. Students have used the converse of the
Pythagorean theorem to show a triangle is a right triangle. This means that we can also take the
converse of CevaÕs Theorem so now we have:

ÒIf ,
then the point M is a point of concurrencyÓ

In order to
prove that this is true we must demonstrate to student what a proof by
contradiction is or an indirect proof.

**Proof of the Converse
of CevaÕs Theorem**

We are given
that .

Assume that
BF and CE intersect at point M, and the other Cevian through M is AD. Thus by CevaÕs theorem

But we had
before that

So by the
transitive property we have that

When we
simplify this we have that

The only way
that this would be true is if H and D are the same point. Thus the Cevians are all intersecting
at the same point.

Having gone
through all of these proofs what is it that we want the students to come away
with. Using the Investigative GSP
file from above we can show the students that thanks CevaÕs Theorem and the
Converse of CevaÕs Theorem we have shown that there is an infinite number of
points of concurrency within a triangle.
In addition, we have shown with our second case that there are an
infinite number of points of concurrency on the exterior of the triangle. So now let us apply this to a more
specific point of concurrency, which is a good application for Math 3 students.

In math 3
students begin to learn about hyperbolas and the formula them. We can show the students that using a
triangle we can construct a hyperbola not just by the formula. Using GSP we can construct a triangle
with similar isosceles triangles on each side:

1. Construct a circle with center G and
two radii GH and HI.

2. Next construct triangle ABC

3. Using point C as a point of rotation
mark angle HGI and rotate line CA by the marked angle. Construct the midpoint of line segment AC
and construct a perpendicular line through the midpoint AC.

4. Find the point of intersection of the
perpendicular line and the rotated line.
From this point construct a line segment to point A to create the
isosceles triangle.

5. Repeat this process for the remaining
sides of the triangle to have three similar isosceles triangles.

6. At this point construct segments PA,
BO, NC. They are three cevians
thus they intersect at point M.

One may explore using this GSP:

Similar Isosceles Triangles

If we push the animate button we see that the trace of the point M
creates a hyperbola. However, we
can show that the asymptotes of this hyperbola are perpendicular to one another
thus creating what is known as a rectangular hyperbola. From Wells __The Penguin Dictionary of
Curious and Interesting Geometry__ we find that ÒIf the three vertices of a
Triangle ABC lie on a rectangular hyperbola, then so does the orthocenter H.
Equivalently, if four points form an orthocentric system, then there is a
family of rectangular hyperbolas through the points.Ó

Using this information we can construct the orthocenter of the triangle
to see whether or not it lies on the path of the hyperbola. This may seem rather simple and truly
how do we know that the orthocenter lies on the locus of point M. The equivalent statement seems more
like a possibility to demonstrate. The definition from orthocentric system from Wolfram Math:

A set of four points, one of which is
the orthocenter of
the other three. In an orthocentric system, each point is the orthocenter of
the triangle of
the other three, as illustrated above (Coxeter and Greitzer 1967, p. 39).

So what we must do is construct the
orthocenter of the triangle and using the three new triangles find that the orthocenter
of those inner triangles are the points of the original triangle.

Given Triangle ABC with orthocenter O

We begin by looking at triangle OAC and
finding the orthocenter of that triangle:

We find the orthocenter to be point B
our original triangle. Now we move
on and find the orthocenter of triangle BOA:

We find that the orthocenter of
Triangle BOA is the point C of our original triangle. Now to finish seeing if this triangle is an orthocentric
system we look at triangle BOC and find its orthocenter:

Now we find that the orthocenter for
Triangle BOC is point A of our original triangle. Thus points A, B, C, and O create an orthocentric
system. Since the triangle is an
orthocentric system then the hyperbola created by tracing point M is a
rectangular hyperbola.