The Nagel Point Theorem

The excircle opposite each vertex touches the opposite side at a point. The lines from each vertex to that tangent point are concurrent at the Nagel point, proven via Ceva's theorem.

Keywords: Nagel point, excircle, tangent point, Ceva's theorem, concurrency

Prerequisites: cevas-theorem-proof, excircles-intro · Difficulty: advanced

The excircle opposite each vertex touches the opposite side at one point: $D^{'}$ on $B C$ , $E^{'}$ on $C A$ , $F^{'}$ on $A B$ . The cevians $A D^{'}$ , $B E^{'}$ , $C F^{'}$ to these tangent points are concurrent, at the Nagel point $N a$ .

Strategy: apply Ceva's theorem. The excircle tangent lengths give a product of side ratios that telescopes to $1$ , forcing concurrency.

Step 1: Excircles and Their Tangent Points

Every triangle has three excircles, one opposite each vertex. The excircle opposite $A$ is tangent to side $B C$ (and to the extensions of $A B$ and $C A$ ); its tangent point on $B C$ is $D^{'}$ .

Define $E^{'}$ on $C A$ and $F^{'}$ on $A B$ analogously, as the tangent points of the excircles opposite $B$ and $C$ .

Step 2: Cevians to Tangent Points

Draw the cevian from each vertex to the excircle tangent point on the opposite side: $A D^{'}$ , $B E^{'}$ , $C F^{'}$ . We claim these three cevians are concurrent.

Step 3: The Nagel Point Theorem

The Nagel Point Theorem: the three cevians $A D^{'}$ , $B E^{'}$ , $C F^{'}$ from each vertex to the opposite excircle tangent point are concurrent. Their common point is the Nagel point $N a$ .

We'll prove it using Ceva's Theorem — the equal-tangent- length property of each excircle makes the Ceva product telescope to $1$ .

Step 4: Tangent Lengths and Ceva's Product

Let $a = B C$ , $b = C A$ , $c = A B$ and $s = (a + b + c) /2$ be the semi-perimeter. The two tangent segments from each vertex to a given excircle are equal, which yields:

$B D^{'} = s - c$ , $D^{'} C = s - b$
$C E^{'} = s - a$ , $E^{'} A = s - c$
$A F^{'} = s - b$ , $F^{'} B = s - a$

(Note these are "reversed" compared to the incircle tangent lengths: the excircle opposite $A$ gives $B D^{'} = s - c$ rather than $s - b$ .)

Ceva's product becomes:

\frac{B D ^{'}}{D ^{'} C} \cdot \frac{C E ^{'}}{E ^{'} A} \cdot \frac{A F ^{'}}{F ^{'} B} = \frac{s - c}{s - b} \cdot \frac{s - a}{s - c} \cdot \frac{s - b}{s - a} = 1

Every factor cancels, so by Ceva's theorem the cevians are concurrent.

Step 5: The Nagel Point

By Ceva's theorem, the three cevians $A D^{'}$ , $B E^{'}$ , $C F^{'}$ are concurrent. Their common intersection is the Nagel point $N a$ .

The Nagel point is the isotomic conjugate of the incenter. It lies on the line from the incenter $I$ through the centroid $G$ , with $G$ dividing $I N a$ in the ratio $1 : 2$ .

The lines from each vertex to the tangent point of the opposite excircle are concurrent at the Nagel point $N a$ .

Like the Gergonne point, this follows from Ceva's theorem: the excircle tangent lengths yield a telescoping product equal to $1$ .

The Nagel point is the isotomic conjugate of the incenter, and lies on the line from the incenter through the centroid (in ratio $1 : 2$ ). ∎