## Final Assignment: Ceva's Theorem

### by: Lauren Wright

For this assignment, we consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively.

Then, we explore (AF)(BD)(CE) and (FB)(DC)(EA) for various triangles and various locations of P.

After exploring this in GSP, I came to the conclusion that (AF)(BD)(CE) = (FB)(DC)(EA) whenever P is inside the triangle. Or, more simply stated, we have:

This is Ceva's Theorem, which states:

In a triangle ABC, three lines AD, BE and CF intersect at a single point P if and only if

(The lines that meet at a point are said to be concurrent.)

Now for a proof...

Extend the lines CF and BE beyond the triangle until they meet GH, the line through A parallel to BC.

There are several pairs of similar triangles: AHE and CBE, AFG and CBF, AGP and CDP, BDP and AHP. From these and in that order we derive the following proportions:

from the last two we conclude that

.

Therefore,

.

Multiplying #1, #2, and #6 together, we get:

Therefore, if the lines AD, BE and CF intersect at a single point P, the identity does hold. Which is to say that the fact of the three lines intersecting at one point is sufficient for the condition to hold.

Let us now prove that it's also necessary. This would constitute the second part of the theorem. In other words, let us prove that if holds then AD, BE, CF are concurrent.

Let's assume that P is the point of intersection of BE and CF and draw the line AP until its intersection with BC at a point D*. Then, from the just proven part of the theorem it follows that

And it's given that

Combining these two, we get the following:

Therefore,

which implies D*C=DC. Or, D* and D are one and the same point.

Click here to see the website that assisted me in this proof.

This result can be generalized to (using lines rather than segments to construct ABC) so that point P can be outside the triangle. Click here for a working GSP sketch.

For the last exploration, we will look at the ratio of the areas of triangle ABC and triangle DEF when P is inside triangle ABC.

After building and exploring this triangle, I found that the ratio of the areas is always greater than or equal to 4. First, I want to look at the case when the ratio is equal to 4.

I noticed that D, E, and F appear to be the midpoints of the sides of ABC when the ratio is equal to 4. I decided that this must mean that the ratio is equal to 4 when P is the centroid of ABC. That means that DEF is the medial triangle. I know that the medial triangle is exactly one fourth of the original triangle, so this makes sense for me. Now, if P is not the centroid, why is the ratio of the areas going to be greater than 4?

Well, I know that the medial triangle is the largest triangle that can be formed inside the original triangle, so the ratio of 4 must be the smallest that the ratio of ABC to DEF can be - or greater than or equal to 4!