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 Ceva’s 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 Ceva’s Theorem is true; what is proof for that?
What our goal is with this proof for Ceva’s 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 Ceva’s Theorem
Given: Triangle ABC with Cevians AD, BF, and CE with M being the point of concurrency
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 Ceva’s theorem still holds. So we must now demonstrate that Ceva’s Theorem must also be proven for this case, showing students that some proofs require us to look at multiple cases or scenarios.
Continuation of Ceva’s Theorem
Given: Triangle ABC with Cevians AD, BF, and CE with M being the point of concurrency on the outside of the triangle
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 Ceva’s Theorem again:
“If the point M is a point of concurrency then . “
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 Ceva’s Theorem so now we have:
“If , then the point M is a point of concurrency”
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 Ceva’s Theorem
We are given that .
Assume that BF and CE intersect at point M, and the other Cevian through M is AD. Thus by Ceva’s 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 Ceva’s Theorem and the Converse of Ceva’s 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 “If 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.”
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.