# Incircles and Excircles in a Triangle

The points of tangency of the incircle of $\Delta ABC$ with sides $a,$ $b,$ $c,$ and semiperimeter $p = (a + b + c)/2,$ define the cevians that meet at the Gergonne point of the triangle. This follows immediately from Ceva's theorem and the fact that two tangents to a circle from a point outside the circle are equal.

The length of those tangents from the vertices of the triangle to its incircle can be easily determined. Denote them $x,$ $y,$ $z,$ as in the diagram. We have three equations:

(1) | $\begin{align} x + y &= c \\ y + z &= a \\ z + x &= b, \end{align}$ |

from where

(2) | $x + y + z = p.$ |

Subtracting from (2) equations (1) one at a time, we get

(3) | $\begin{align} x &= p - a\\ y &= p - b \\ z &= p - c. \end{align}$ |

These are the lengths that appear in Heron's formula, e.g., $p - a = (b + c - a)/2.$

Similarly we can find the lengths of the tangents to the excircles.

Obviously, $2p = (b + u) + (c + v).$ But, since $(b + u) = (c + v),$ we get

(4) | $\begin{align} u &= p - b \\ v &= p - c. \end{align}$ |

From the just derived formulas it follows that the points of tangency of the incircle and an excircle with a side of a triangle are symmetric with respect to the midpoint of the side. Such points are called *isotomic*. The cevians joinging the two points to the opposite vertex are also said to be isotomic. Both triples of cevians meet in a point. For the incircle, the point is Gergonne'; for the points of excircle tangency, the point is Nagel's. We have just proved that, in any triangle, the Gergonne and Nagel points are *isotomic conjugate* of each other. (This fact has an interesting geometric illustration.)

In general, two points in a triangle are *isotomic conjugate* if the cevians through them are pairwise isotomic. The centroid is one point that is its own isotomic conjugate.

### cot(A/2) = (p - a)/r

This obvious formula sometimes goes under the name of *The Law of Cotangents*:

$\displaystyle\frac{\cot (A/2)}{p-a}=\frac{\cot (B/2)}{p-b}=\frac{\cot (C/2)}{p-c}=\frac{1}{r}.$

### S = r_{a}(p - a)

Indeed,

$\begin{align} 2S &= (b + u)r_{a} + (c + v)r_{a} - ar_{a} - (u + v)r_{a} \\ &= (b + c - a)r_{a} \\ &= (2p - 2a)r_{a}\\ &= 2(p - a)r_{a}. \end{align}$

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