** **

**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. (1) Explore (AF)(BD)(EC) and
(FB)(DC)(EA) for various triangles and various locations of P; (2) Show that
when P is inside ****D****ABC, the ratio of the
areas of ****D****ABC and ****D****DEF is always greater
than or equal to 4..**

Here’s
a sketch:

Click
here for a sketch to experiment with.
You’ll see that the ratio of the segments is always 1 when P is
inside or outside the triangle.
When P is actually on one of the segments of the triangle, the ratio is
either zero or undefined. The only
way I could see that precisely with GSP was to merge P with one of the segments
of the triangle (under the Edit menu, if you’re interested.) Another interesting case is when the
triangle is degenerate, that is the three vertices are collinear. Then …

This
phenomenon is known as Ceva’s theorem, which states that three lines AD,
BE and CF intersect in a single point if and only if (AF/FB)*(BD/DC)*(CD/EA) =
1. The lines AD, CF and BE are
known as *Cevians,*
lines in a triangle from a vertex to the opposite side.

Why
does it work? To prove, add a
construction line to form similar triangles. Creating a line parallel to BC through A will do it. Let the intersection of the line with
extended CF be Q, and the intersection of the line with extended BE be R. This creates lots of “sets”
of similar triangles. Here’s
a sketch of the first:

The
yellow triangles, DAQF and DBFC are similar (use QC
and AB as transversals of parallel lines QR and BC to see this.) Likewise, the gray triangles, DAER and DCEB are similar.

Notice
that DAPQ is similar to DDPC (the pink triangles)
and that the purple triangles, DAPR and DDPB are similar.

·
From
the yellow triangles: **(AF/FB)** = (FQ/FC) = **(AQ/BC).**

·
From
the gray triangles: **(EC/AE)** = (BE/RE) = **(BC/AR).**

·
From
the pink triangles: (AP/PD) =
(QP/PC) = (AQ/DC), and

·
From
the purple triangles: (AP/PD) =
(RP/PB) = (AR/BD).

The
ratios in the pink and purple triangles are all equal, since they have (AP/PD)
in common, so we can set (AQ/DC) = (AR/BD). A little algebra gives **(BD/DC)** = **(AR/AQ)**.

Next,
multiply together the bolded terms to get:

(AF/FB)
(EC/AE) (BD/DC) = (AQ/BC) (BC/AR) (AR/AQ)

The
terms on the right side all cancel out, giving us

(AF/FB) (EC/AE) (BD/DC) = 1!

Here’s
a sketch with the area ratio:

This
sketch has the common triangle centers added as well. If you move P over the centroid, the area ratio is always
4. If P is anywhere else within
the triangle, the ratio is greater than 4. On this sketch, you can also move P outside the
triangle.

So
we ask the usual question: why
does this work? When P is the
centroid, the Cevians are the medians, and D, E and F are the midpoints of the
sides of DABC, and the triangle mid-segment theorem comes to
the rescue. By that theorem, each
of the white triangles (DBDF, DAFE, and DDEC) have a base that is 1/2
the base of DABC and a height that is 1/2 the height of DABC. That gives each of the white triangles
an area that is 1/4 the area of DABC, leaving 1/4 of the
area in DDEF.

Some
interesting extensions of this exploration would be to use Ceva’s theorem
to prove that the lines that create the common triangle centers (orthocenter,
incenter, circumcenter and centroid) are concurrent. I think this is a great exploration for kids, because
Ceva’s theorem is pretty obvious using the measurements on GSP. It’s also an easy proof, once you
draw in the right construction line.
Have fun with it!