Final Assignment
Ceva's Theorem
by Michael Ferra
Proposed Investigation
Given a △ABC with an arbitrary point P inside the triangle. Any segment from a vertex to the intersection with the opposite side is called a Cevian. Let D, E and F be the feet of the Cevians opposite A, B and C respectively.
Ceva's Theorem: If the three Cevians AD, BE and CF are concurrent at P, then
.
Converse of Ceva's Theorem: If
i. Prove Ceva's Theorem using ratios of areas.
ii. Prove the converse of Ceva's Theorem.
iii. Use the converse of Ceva's Theorem to prove the medians of a triangle are concurrent.
i. Prove Ceva's Theorem using ratios of areas.
Let's begin by constructing any △ABC, an arbitrary point P inside △ABC, and the resulting Cevians AD, BE and CF of △ABC.
If the three Cevians AD, BE and CF are concurrent at P, prove
Observe △ABD and △ACD have sides BD and CD that both lie on segment BC of △ABC. Also notice the vertex opposite of BD in △ABD is the same vertex that is opposite of CD in △ACD, which is the shared vertex A. What is helpful about this?
When finding the area of these two triangles, we will use BD as the base for △ABD and CD as the base for △ACD. We still need to find the altitude of these triangles to find their areas, so to define the altitude we construct a line perpendicular to the base through its opposite vertex. Since BD and CD lie on the same line and since they share the same opposite vertex, their altitude will be the same!! Let's construct this altitude and call it AG.
Let's make a similar observation as above to notice △BCE and △ABE have sides CE and AE that both lie on segment AC of △ABC. Also notice the vertex opposite of CE in △BCE is the same vertex that is opposite of AE in △ABE, which is the shared vertex B.
When finding the area of these two triangles, we will use CE as the base for △BCE and AE as the base for △ABE. Notice the same idea before that CE and AE lie on the same line, and since they share the same opposite vertex, their altitude will be the same!! Let's construct this altitude and call it BH.
▩
ii. Prove the converse of Ceva's Theorem.
If
Suppose the Cevians AD and BE met at our arbitrary point P. The third cevian through this point P is CK. By the previous proof, we have
, but we have assumed
. Therefore K = F, thus AD, BE and CF are concurrent. ▩
iii. Use the converse of Ceva's Theorem to prove the medians of a triangle are concurrent.
Given △ABC, define the point P inside the triangle to be the centroid. Cevians AD, BE and CF are then the medians of △ABC. Therefore BD = CD, CE = AE, and AF = BF, thus (BD)(CE)(AF) = (CD)(AE)(BF). Since (BD)(CE)(AF) = (CD)(AE)(BF) we then know
