Alfred S. Posamentier - The Joy of Mathematics: Marvels, Novelties, and Neglected Gems That Are Rarely Taught in Math Class

As we mentioned earlier, Ceva's theorem might well have been introduced to a high school class, since it merely applies similarity relationships that are an integral part of the geometry curriculum. We offer one of many proofs available to justify Ceva's theorem. It is perhaps easier to follow the proof by looking at the left-side diagram in and then verifying the validity of each of the statements in the right-side diagram. In any case, the statements made in the proof hold for bothdiagrams.

Figure App.1.

Consider , for which we have on the left triangle ABCwith a line ( SR) containing Aand parallel to BC, intersecting CPextended at Sand BPextended at R.

The parallel lines enable us to establish the following pairs of similar triangles:

Now by multiplying (I), (II), and (V), we obtain our desired result:

This can also be written as AMBNCL= MCNABL. A nice way to read this theorem is that the product of the alternate segments along the sides of the triangle made by the concurrent line segments (called cevians) emanating from the triangle's vertices and ending at the opposite side are equal.

Yet, it is the converse of this proof that is of particular value to use here. That is, if the products of the alternate segments along the sides of the triangle are equal, then the cevians determining these points must be concurrent.

We shall now prove that if the lines containing the vertices of triangle ABCintersect the opposite sides in points L, M, and N, respectively, so that , then these lines, AL, BM, and CN, are concurrent.

Suppose BMand ALintersect at P. Draw PCand call its intersection with