(by Polymath) This page is part of the Advanced High School Math Project.

If
you have learned anything about triangle centers from a basic high
school geometry course, you have almost surely learned about the four classical centers.
But in addition to those, there are two other triangle centers that are
fairly easily established with basic high school math, especially once
you've mastered the use of Ceva's theorem. Those are the Gergonne
point and the Nagel point, and I will prove their existence in this
post. It is worth noting that, basic as these proofs and ideas are,
they were not discovered until early in the 19th century; not much
before, say, Riemann posed his famous hypothesis, drawing on a *much* deeper set of mathematical principles.

But on to the proofs. The reason these two triangle centers belong together in one post is that they both involve cevians drawn from a vertex to the points of tangency of circlesâ€”and closely related circles, at that. The Gergonne point is the concurrence point for the cevians from each vertex to the point on the opposite side where the inscribed circle is tangent. The Nagel point is the concurrence point for the cevians from each vertex to the point on the opposite side where that side's escribed circle is tangent (the escribed circles, or excircles are lesser known than the inscribed circle, and I will go into more detail about them below). You can see just from the similar descriptions of the points that they must be closely related.

The cevians in red in the triangle below connect, as described above, each vertex of the triangle with the point of tangency of the inscribed circle. Note that since two tangent segments to the same circle from a single point are congruent, I've marked three pairs of congruent segments with $, *, and #.

Recall that Ceva's theorem says that the three segments concur if

But now, this should be obvious: replacing each segment name with its equivalent $, #, or * puts one of each of those symbols in the numerator and one in the denominator, making the product equal 1, just as desired. Thus the segments concur. And we call that point the Gergonne point.

The Nagel point is a
little trickier, but also a little more interesting. It requires some
deeper understanding of excircles, so let's start with those. You
already know (you don't? click here)
that an inscribed circle is tangent to all three sides of a triangle
because its center is equidistant from the three sides. Furthermore,
that equidistance is generated by the fact that the center lies on all
three angle bisectors, which are rays consisting of all the points that
are equidistant from the sides. But more advanced views of triangle
geometry require that we look not just *inside* a triangle, but *outside it*
as well. If you allow the idea that a side of a triangle is not merely
a segment, but the whole line containing the segment, then the center
of the inscribed circle is not the only point equidistant from the
sides of the triangle. Consider the angle bisector of an exterior
angle, for instance. It consists of all the points equidistant from
one side of a triangle and the extension of another side. Drawing the
bisector of the nearby exterior angle (if you used the extension of AB
the first time, then use the extension of AC for the nearby angle),
then, creates an intersection point that is equidistant from all three
sides just as surely as drawing
two interior angle bisectors does. That intersection point outside the
triangle (which, of course, must lie on the bisector of the hitherto
unused third interior angle, A in our example) is the center of one of
three excircles. Below is a picture of a triangle and all three of its
excircles.

Just as the incircle is tangent to all three sides, you'll note that each excircle is tangent to all three sides if, again, you relax your understanding of "sides" to include the lines and not just the segments that make up the sides.

But...the segments are, of course, still the primary components of the triangle. And each of the side segments of the original triangle does indeed contain one point of tangency for each excircle. The Nagel point, then, is the concurrence point of the three cevians connecting each vertex of the triangle to that point of tangency of the corresponding excircle on the opposite side.

The proof that those cevians concur requires that we first see those points of tangency in a different light. Below is an enlargement of the middle of the picture above, showing the point of tangency (X) of the excircle opposite point A, along with the other points of tangency of that excircle (P and Q) on the extensions of the sides.

Again, we use the theorem saying that two tangent segments to a single
circle from the same point are congruent: thus the two green segments
form a congruent pair, as do the two blue and the two purple segments.
Now we take a walk. First walk from A to B to X. Because of the blue
segments, that walk is the same length as the walk from A to B to Q.
Because of the green segments, *that* walk is the same length as the walk from A to C to P. And because of the purple segments, * that* walk is the same length as the walk from A to C to X. Thus the segment sum AB+BX equals the segment sum AC+CX.

So the point X can now be characterized in two different ways. First, it is the point of tangency of the excircle opposite A. Second, it marks the point halfway around the perimeter in either direction from A (the semiperimeter point from A).

The length of the semiperimeter is commonly represented by *s*. Using the standard triangle labeling system, where side *c* is opposite angle C (thus, the length of AB), we can now represent the length of BX with (*s* - *c*).
That is, the combined lengths of AB and BX make the semiperimeter, and
we get an expression for BX by subtracting the length of AB. For
identical reasons, we can devise similar expressions for the other
segments shown below (Y and Z are, of course, the two other points of excircle
tangency).

But now the proof is easy. Ceva's equation clearly holds:

So those cevians indeed concur, and we call that point the Nagel point of the triangle.