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Ceva’s Theorem

By Mary Negley

For this write-up, I will explore Ceva’s Theorem, which states that given a triangle , the three lines , , and intersect at point iff .  A GSP sketch is given below. An interesting fact about point P is that it may be anywhere on the plane. I will look at P in the following cases: (1) when P is the centroid, (2) when P is the orthocenter, (3) when P is the incenter, and (4) when P is the circumcenter.

Case1: P is the centroid.  By definition of centroid, P is the intersection of the medians of the triangle.  Below is a GSP sketch of this case. Clearly, the ratio holds in this case.

Case2: P is the orthocenter.  By definition of orthocenter, P is the intersection of the lines of the altitudes of the triangle.  Below is a GSP sketch of this case. Clearly, the ratio holds in this case.

Case3: P is the incenter.  By definition of incenter, P is the intersection of the angle bisectors of the triangle.  Below is a GSP sketch of this case. Clearly, the ratio holds in this case.

Case4: P is the circumcenter.  By definition of circumcenter, P is the intersection of the perpendicular bisectors of the sides of the triangle.  Below is a GSP sketch of this case. Clearly, the ratio holds in this case.

Proof:  Now I will show that given a triangle , the three lines , , and intersect at point iff .

First I will assume that the three lines intersect at point P and then I will show that .

Below is the triangle. To aid me in my proof, I will extend the segments AD, BE, and CF to be lines and I will draw a line through C, which is parallel to AB.  Below is the new sketch. Now I will shade in a couple of the similar triangles that occur. Triangle GCP ~ Triangle AFP because

1) by congruency of alternate interior angles

2) by congruency of vertical angle theorem

and    3) since and , and by 1) and 2) so subtracting <PGC and <CPG from both sides yields .

Similarly, Triangle GCD ~ Triangle ABD, Triangle HCP ~ Triangle BFP and Triangle HEC ~ Triangle BEA,

Since the preceding triangles are similar, the following ratios occur:

(1) (2) (3) (4) To show that , I can show that .

By using (1), (3) and the property of transitivity, I get that .  Solving for HC, I get .  Solving for HC in (4) yields .  By transitivity, .  Now I will cross-multiply and get .  Then, solving for GC in (2) yields and when I substitute that into my equality, I get .  By simplifying and cross-multiplying, I get .  Hence, .

Thus far, I have shown one part of the proof. Now I will show the other part of the proof.  I must assume that and then show that P must be a point of concurrence.

Below is a GSP sketch of the triangle that is given. Assume that BE and CF intersect at P and AX has been constructed to go through P, but .

Since AD and AX both go through P, we know that and .

Then, by transitivity, .

After simplifying, I get .

So .

Hence, P is the point of concurrence of , , and .  Therefore, I have proved the theorem.

A question that arises is how might this theorem help us to prove concurrency of the lines of altitudes if P is the orthocenter?

First we need a sketch of when P is the orthocenter. The theorem states that given a triangle , the three lines , , and intersect at point iff .

Let P be the orthocenter.  Then, by definition of orthocenter, the lines through P from each vertex must be perpendicular to the sides of the triangle.  Notice that Triangle ABD is similar to Triangle CBF by angle-angle similarity (they share the angle B and they each have a right angle by construction).  Similarly (ha ha!), Triangle ACD is similar to Triangle BCE and Triangle BAE is similar to Triangle CAF.  So the following proportions result: ,  and By Ceva’s Theorem, the three lines , , and intersect at point iff .  Let’s assume that .  Substituting values from above, gives us .  Therefore, the altitudes must be concurrent.

Next I will generalize my result so that the point P may be outside the triangle.  In order to do that, I must extend the sides of the triangle to be lines.  A working sketch of this case may be found here. 