Incircles and Excircles

Introduction

How would you draw a circle inside a triangle, touching all three sides? It is actually not too complex. Simply bisect each of the angles of the triangle; the point where they meet is the center of the circle! Then use a compass to draw the circle. But what else did you discover doing this?

1. The three angle bisectors all meet at one point.
2. This point is equidistant from all three sides.

In order to prove these statements and to explore further, we establish some notation.

Let $AU$, $BV$ and $CW$ be the angle bisectors.
The incenter $I$ is the point where the angle bisectors meet.

Let $X, Y$ and $Z$ be the perpendiculars from the incenter to each of the sides.
The incircle is the inscribed circle of the triangle that touches all three sides.

The inradius $r$ is the radius of the incircle.

Now we prove the statements discovered in the introduction.

In a triangle $ABC$, the angle bisectors of the three angles are concurrent at the incenter $I$. Also, the incenter is the center of the incircle inscribed in the triangle.

Given $\triangle ABC,$ place point $U$ on $\overline{BC}$ such that $\overline{AU}$ bisects $\angle A,$ and place point $V$ on $\overline{AC}$ such that $\overline{BV}$ bisects $\angle B.$ Let $I$ be their point of intersection. Then place point $X$ on $\overline{BC}$ such that $\overline{IX} \perp \overline{BC},$ place point $Y$ on $\overline{AC}$ such that $\overline{IY} \perp \overline{AC},$ and place point $Z$ on $\overline{AB}$ such that $\overline{IZ} \perp \overline{AB}.$ Finally, place point $W$ on $\overline{AB}$ such that $\overline{CW}$ passes through point $I.$

$\triangle AIY$ and $\triangle AIZ$ have the following congruences:

• $\angle AYI \cong \angle AZI$ because they are both right angles.
• $\angle IAY \cong \angle IAZ$ because $\overline{AI}$ is the angle bisector.
• $\overline{AI} \cong \overline{AI}$ because of the reflexive property of congruence.

Thus, by AAS, $\triangle AIY \cong \triangle AIZ.$ In a similar fashion, it can be proven that $\triangle BIX \cong \triangle BIZ.$ Then, by CPCTC (congruent parts of congruent triangles are congruent) and the transitive property of congruence, $\overline{IX} \cong \overline{IY} \cong \overline{IZ}.$

Now $\triangle CIX$ and $\triangle CIY$ have the following congruences:

• $\overline{IX} \cong \overline{IY},$ as stated earlier.
• $\angle CXI \cong \angle CYI$ because they are both right angles.
• $\overline{CI} \cong \overline{CI}$ because of the reflexive property of congruence.

Thus, by HL (hypotenuse-leg theorem), $\triangle CIX \cong \triangle CIY.$ By CPCTC, $\angle ICX \cong \angle ICY.$ Hence, $\overline{CW}$ is the angle bisector of $\angle C,$ and all three angle bisectors meet at point $I.$

Since $\overline{IX} \cong \overline{IY} \cong \overline{IZ},$ there exists a circle centered at $I$ that passes through $X,$ $Y,$ and $Z.$ Furthermore, since these segments are perpendicular to the sides of the triangle, the circle is internally tangent to the triangle at each of these points. Hence, the incenter is located at point $I.$ $_\square$

If we extend two of the sides of the triangle, we can get a similar configuration.

Note that these notations cycle for all three ways to extend two sides $(A1, B2, C3).$ $I_1$ is the excenter opposite $A$. It has two main properties:

1. The angle bisectors of $\angle A, \angle Z_1BC, \angle Y_1CB$ are all concurrent at $I_1$.
2. $I_1$ is the center of the excircle which is the circle tangent to $BC$ and to the extensions of $AB$ and $AC$.

$r_1$ is the radius of the excircle.

The proofs of these results are very similar to those with incircles, so they are left to the reader.

There are many amazing properties of these configurations, but here are the main ones. In these theorems the semi-perimeter $s = \frac{a+b+c}{2}$, and the area of a triangle $XYZ$ is denoted $\left[XYZ\right]$.

Elementary Length Formulae:

First we prove two similar theorems related to lengths.

$$AY = AZ = s-a,\quad BZ = BX = s-b,\quad CX = CY = s-c.$$

Tangents from the same point are equal, so $AY = AZ$ (and cyclic results). Then it follows that $AY + BW + CX = s$, but $BW = BX$, so

$$\begin{aligned} AY + BX + CX &= s \\ AY + a &=s \\ AY &= s-a, \end{aligned}$$

and the result follows immediately. $_\square$

The argument is very similar for the other two results, so it is left to the reader.

$$BX_1 = BZ_1 = s-c,\quad CY_1 = CX_1 = s-b,\quad AY_1 = AZ_1 = s.$$

The proof of this theorem is quite similar and is left to the reader.

Area Formulae:

This is a beautiful theorem about areas:

$$\left[ABC\right] = rs = r_1(s-a) = r_2(s-b) = r_3(s-c)$$

The proof is left to the reader for now.

These more advanced, but useful properties will be listed for the reader to prove (as exercises).

Radii Relationships:

These are very useful when dealing with problems involving the inradius and the exradii. $($Let $R$ be the circumradius.$)$

$$\begin{aligned} \frac{1}{r} &= \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\\\\ r_1 + r_2 + r_3 - r &= 4R \\\\ s^2 &= r_1r_2 + r_2r_3 + r_3r_1. \end{aligned}$$

And here is one of my favorites:

$$\left[ ABC\right] = \sqrt{rr_1r_2r_3}.$$

Computing Lengths:

$$\begin{aligned} AI &= r\mathrm{cosec} \left({\frac{1}{2}A}\right) \\\\ r &= \sqrt{\frac{(s-a)(s-b)(s-c)}{s}} \end{aligned}$$