The Gergonne Point Theorem

The incircle touches each side at a point. The lines from each vertex to the opposite tangent point are concurrent at the Gergonne point, proven via Ceva's theorem and tangent length ratios.

Keywords: Gergonne point, incircle, tangent point, Ceva's theorem, concurrency

Prerequisites: cevas-theorem-proof, incenter-proof · Difficulty: intermediate

The incircle of $△ A B C$ touches side $B C$ at $D$ , side $C A$ at $E$ , and side $A B$ at $F$ . The three lines $A D$ , $B E$ , $C F$ are concurrent at a point called the Gergonne point $G e$ .

Strategy: apply Ceva's theorem. The equal tangent lengths from the incircle make the Ceva product telescope to $1$ , so the three cevians meet at one point.

Step 1: Incircle and Tangent Points

Begin with triangle $A B C$ . Construct the incircle, which touches each side at exactly one point: $D$ on $B C$ , $E$ on $C A$ , $F$ on $A B$ .

Step 2: Cevians to Tangent Points

Draw the line from each vertex to the tangent point on the opposite side: $A D$ , $B E$ , and $C F$ . We claim these three cevians are concurrent.

Step 3: The Gergonne Point Theorem

The Gergonne Point Theorem: the three cevians $A D$ , $B E$ , $C F$ from each vertex to the opposite incircle tangent point are concurrent. Their common point is the Gergonne point $G e$ .

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

Step 4: Tangent Ratios: Product $= 1$

Let $a = B C$ , $b = C A$ , $c = A B$ be the side lengths opposite each vertex, and let $s = (a + b + c) /2$ be the semi-perimeter.

The two tangent segments from each vertex to the incircle are equal, giving:

$B D = s - b$ , $D C = s - c$
$C E = s - c$ , $E A = s - a$
$A F = s - a$ , $F B = s - b$

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

By Ceva's theorem the cevians are concurrent.

Step 5: The Gergonne Point

By Ceva's theorem, since the product of the ratios is $1$ , the three cevians $A D$ , $B E$ , $C F$ are concurrent. Their common point is the Gergonne point $G e$ .

The lines from each vertex to the point where the incircle touches the opposite side are concurrent at the Gergonne point $G e$ .

The proof is a direct application of Ceva's theorem: the tangent lengths from the incircle yield ratios whose product telescopes to $1$ . ∎

Notes

Notation

Let $a = B C$ , $b = C A$ , $c = A B$ be the side lengths opposite the vertices $A$ , $B$ , $C$ , and let

s = \frac{a + b + c}{2}

denote the semi-perimeter of the triangle.

Tangent Lengths

The two tangent segments from an external point to a circle have equal length. Applying this at each vertex of $△ A B C$ and solving the resulting three linear equations gives:

From $B$ : $B D = B F = s - b$
From $C$ : $C D = C E = s - c$
From $A$ : $A E = A F = s - a$