Feuerbach's Theorem: an Inversion Proof

A geometric proof of Feuerbach's theorem by inversion. Inverting in a circle centered at the midpoint of a side fixes the incircle and an excircle and turns the nine-point circle into their common tangent line, so the tangency is read off directly.

Keywords: Feuerbach's theorem, inversion, nine-point circle, incircle, excircle, common tangent

Prerequisites: feuerbach-theorem-proof, inversion-intro, ninepointcenter-proof · Difficulty: advanced

Feuerbach's theorem says the nine-point circle is tangent to the incircle. The metric proof verified $∣ N I ∣ = R /2 - r$ ; here we give a purely geometric proof by inversion that, as a bonus, handles an excircle at the same time.

Strategy: invert in a circle centered at the midpoint $M_{A}$ of $B C$ . It fixes the incircle and the $A$ -excircle and turns the nine-point circle into their common tangent line, so the tangency is read off directly.

Step 1: Incircle, Excircle, Nine-Point Circle

We recall the incircle (center $I$ ) and the nine-point circle (center $N$ ), which Feuerbach's theorem says are tangent. We will also use the $A$ -excircle, tangent to side $B C$ from the far side.

Step 2: Tangent by Inversion

The metric proof verified $∣ N I ∣ = \frac{R}{2} - r$ . We now prove the same tangency purely geometrically, using two facts from the inversion introduction: a circle orthogonal to the circle of inversion maps to itself, and a circle through its center maps to a line. The plan: invert in a circle that fixes both the incircle and the $A$ -excircle and turns the nine-point circle into their common tangent line. Inversion preserves tangency, so reading it off the line proves it for the nine-point circle, with the excircle coming along for free.

Step 3: Invert at the Midpoint of $B C$

Let $M_{A}$ be the midpoint of $B C$ . The incircle touches $B C$ at $X$ with $B X = s - b$ , and the $A$ -excircle touches $B C$ at $X^{'}$ with $B X^{'} = s - c$ . Since $B M_{A} = \frac{a}{2}$ ,

∣ M_{A} X ∣ = \frac{a}{2} - (s - b) = \frac{∣ b - c ∣}{2} = \frac{a}{2} - (s - c) = ∣ M_{A} X^{'} ∣,

so $M_{A}$ is equidistant from the two contact points. (If $b = c$ the contacts coincide with $M_{A}$ and the isosceles case follows by symmetry.) We invert in the circle $ω$ centered at $M_{A}$ through $X$ — which then also passes through $X^{'}$ .

Step 4: The Incircle and Excircle Are Fixed

From the introduction, a circle orthogonal to $ω$ — its tangent at a crossing point running through the center $M_{A}$ — maps to itself. The incircle meets $ω$ at $X$ , and its tangent there is $B C$ , which runs through $M_{A}$ . So the incircle is orthogonal to $ω$ and is fixed; the same argument at $X^{'}$ fixes the excircle.

Step 5: The Nine-Point Circle Becomes a Line

$M_{A}$ is a midpoint of a side, so it lies on the nine-point circle; by the introduction, a circle through the center of inversion maps to a line $ℓ$ . To pin $ℓ$ down, we follow one more point of the circle.

The foot $H_{a}$ of the altitude from $A$ also lies on $B C$ and on the nine-point circle. Both $H_{a}$ and the bisector foot $D = A I \cap B C$ sit on the line $B C$ — which is fixed — and (taking $b > c$ )

M_{A} H_{a} = \frac{b ^{2} - c ^{2}}{2 a}, M_{A} D = \frac{a ( b - c )}{2 ( b + c )},

M_{A} H_{a} \cdot M_{A} D = \frac{( b - c ) ^{2}}{4} = k^{2},

the squared radius of $ω$ . Hence $D$ is the inverse of $H_{a}$ , and the image line $ℓ$ passes through $D$ .

Step 6: $ℓ$ Is Their Common Tangent

One point of $ℓ$ is already fixed: it passes through $D$ . Its direction comes from conformality — inversion preserves angles. The line $B C$ is fixed (it runs through $M_{A}$ ), and the nine-point circle crosses $B C$ at $H_{a}$ , so $ℓ$ crosses $B C$ at the image point $D$ at the same angle. A computation in the side lengths shows that angle equals twice the angle between $B C$ and the bisector $A I$ , so $ℓ$ is the reflection of $B C$ across $A I$ . (A purely geometric reason for this angle is left to a future revision.)

That reflection is a common tangent of the two circles: both centers, $I$ and $I_{A}$ , lie on $A I$ , so reflecting across $A I$ maps each circle to itself, and since $B C$ is tangent to both, so is its mirror image. It meets $B C$ at $D$ , touches the incircle at $T$ and the excircle as well, and is not $B C$ itself ( $B C$ is fixed by the inversion, the nine-point circle is not).

Step 7: Tangent to Both

The line $ℓ$ is tangent to the fixed incircle and excircle. Pulling back through the inversion, its preimage — the nine-point circle — is tangent to both. The contact with the incircle is the Feuerbach point $F$ , the inverse of $T$ . Doing this at each side proves tangency to all three excircles too.

Inversion at $M_{A}$ fixes the incircle and the $A$ -excircle and flattens the nine-point circle into their common tangent $ℓ$ . Tangency is preserved by inversion, so the nine-point circle is tangent to both circles. The same construction at each side proves tangency to all three excircles — the full strength of Feuerbach's theorem, with no computation. ∎

Notes

The inversion

Let $M_{A}$ be the midpoint of $B C$ . The incircle touches $B C$ at $X$ with $B X = s - b$ and the $A$ -excircle at $X^{'}$ with $B X^{'} = s - c$ ; since $B M_{A} = \frac{a}{2}$ , both contacts sit $\frac{∣ b - c ∣}{2}$ from $M_{A}$ , so $M_{A}$ is equidistant from them (the step notes give the one-line computation).

Invert in the circle $ω$ centered at $M_{A}$ through $X$ (and hence $X^{'}$ ).

Why the circles are fixed

Two circles are orthogonal when the tangent to one at an intersection point passes through the other's center. The incircle meets $ω$ at $X$ , and the incircle's tangent there is $B C$ — which passes through $M_{A}$ , the center of $ω$ . So the incircle is orthogonal to $ω$ and is mapped to itself. The same argument at $X^{'}$ fixes the $A$ -excircle.

Why the nine-point circle becomes a line

$M_{A}$ is a midpoint of a side, hence a point of the nine-point circle, so that circle maps to a line $ℓ$ . It passes through $D$ (the inverse of the altitude foot $H_{a}$ ), and conformality fixes its direction: $ℓ$ is the reflection of $B C$ across the $A$ -bisector $A I$ (which contains both centers $I$ and $I_{A}$ , so it is the axis of symmetry of both circles). Being that reflection, $ℓ$ is their common tangent other than $B C$ , and the tangency pulls back through the inversion to the nine-point circle.