# Incircles and Excircles

#### Contents

## Incircles and Incentres

**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 centre of the circle! Then use a compass to draw the circle. But what else did you discover doing this?

- The three angle bisectors all meet at one point.
- 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

incentre\(I\) is the point where the angle bisectors meet.Let \(X, Y\) and \(Z\) be the perpendiculars from the incentre to each of the sides.

The

incircleis 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 incentre \(I\). Also, the incentre is the centre 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 incentre is located at point \(I.\)

## Excircles and Excentres

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

excentreopposite \(A\). It has two main properties:

- The angle bisectors of \(\angle A\), of \(\angle Z_1BC\) and of \(\angle Y_1CB\) are concurrent at \(I_1\).
- \(I_1\) is the centre of the
excirclewhich 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.

## Main Properties and Examples

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 \(\left[XYZ\right]\).

**1. Elementary Length Formulae**

First we group two similar theorems related to lengths.

## Theorem 2: \(AY = AZ = s-a\), \(BZ = BX = s-b\), \(CX = CY = s-c\)

## Theorem 3: \(BX_1 = BZ_1 = s-c\), \(CY_1 = CX_1 = s-b\), \(AY_1 = AZ_1 = s\)

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 \[AY + BX + CX = s \\ AY + a =s \\ AY = s-a\] and the result follows immediately. \( \square\).

Proof of Theorem 2:The argument is very similar for the other two results, so it is left to the reader.

Proof of Theorem 3:This proof is quite similar and is left to the reader.

Prove that the inradius of triangle \(ABC\), right angled at \(C\), is always an integer.

Here is the outlined proof:Pythagoras' Theorem states that \(a^2 + b^2 = c^2\). Therefore either all \(a,b,c\) are even or two or \(a,b,c\) are odd. Either way, \(s\) is always an integer.Now notice that \(r = s-c\) in this case by Theorem 2, and this is an integer so we are done. \(\square\)

[

**2. Area Formulae**

This is a beautiful theorem about areas:

## Theorem 4: \(\left[ABC\right] = rs = r_1(s-a) = r_2(s-b) = r_3(s-c)\)

The proof is left to the reader for now.

## More Advanced Useful Properties

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

**3. Radii Relationships**

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

\[\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\] \[ r_1 + r_2 + r_3 - r = 4R \] \[s^2 = r_1r_2 + r_2r_3 + r_3r_1\] And here is one of my favourites: \[ \left[ ABC\right] = \sqrt{rr_1r_2r_3}\]

**4. Computing lengths**
\[AI = r\mathrm{cosec} ({\frac{1}{2}A}) \]
\[r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\]

**Cite as:**Incircles and Excircles.

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