Exeter Point

Sohail Farhangi 17 February, 2013

The Exeter point - being found in 1986 - is a relatively recent discovery in triangle geometry. Elsewhere we established the following theorem:

Let the incircle $T$, of $\Delta ABC$ be tangent to the sides at $A'$, $B'$, and $C'$ as shown below. For a point $P$ in the plane, let $AP$, $BP$, and $CP$ intersect $T$ at $A_1$, $B_1$, and $C_1$ respectively. Then $AA_1$, $BB_1$, and $CC_1$ concur at a point $P'$. The Exeter point $E=P'$ for $P$ being the center $G$ of a triangle. Below we prove an important property of Exeter point, to boot:

The Exeter point $E$ lies on the Euler line $L$ of $\Delta A'B'C'$.

The applet below serves a dynamic illustration:

Proof

Let $M_A$, $M_B$, $M_C$ be the midpoints of sides $BC$, $CA$, $AB$, respectively. We begin by inverting lines $AA_1$, $BB_1$, and $CC_1$ with respect to $T$. Noting that $AB'$ and $AC'$ are tangent to $T$, we can see that $A$ maps to $M_A$ under our inversion, and the same can be seen for vertices $B$ and $C$. Points $A_1$, $B_1$, and $C_1$ are invariant under the inversion as they lie on $T$. We may also note that the point at infinity on lines $AA_1$, $BB_1$, $CC_1$ map to the incenter $I$ under our inversion. We can now see that lines $AA_1$, $BB_1$, $CC_1$ map to circles $A_{1}M_{A}I$, $B_{1}M_{B}I$, and $C_{1}M_{C}I$ under our inversion, which will be referred to as $O_A$, $O_B$ and $O_C$, respectively.

Noting that the center of $T$ is the circumcenter of $\Delta A'B'C'$, which is on $L$, we know that $E$ lies on $L$ if and only if $E'$, the image of $E$ under our inversion also lies on $L$. From the definition, $O_A$, $O_B$, $O_C$ concur at $E'$. Therefore, we only need to show that $G$ is on the radical axis of $O_A$, $O_B$, $O_C$, which can be done by showing that it has the same power with respect to any $2$ of these $3$ coaxial circles.

For the next step we prove the following

Lemma

$M_{B}M_{C}C_{1}B_{1}$ is a cyclic quadrilateral.

Proof

$C'B'C_{1}B_{1}$ is a cyclic quadrilateral, so $C'B'$ and $B_{1}C_{1}$ are antiparallel lines. Let the circumcircle of $\Delta B_{1}M_{B}C_{1}$ intersect $\Delta A'B'C'$ at $M'_C$. $M_{B}M'_C$ and $B_{1}C_{1}$ are also antiparallel lines, which implies that $M_{B}M'_{C}$ and $B'C'$ are parallel lines, so $M'_{C} = M_C$ as desired.

Returning to the main proposition, we can see that $B_{1}M_{B}$ and $C_{1}M_{B}C$ intersect at $G$ by definition. Applying Lemma tells us that $GM_{B} \times GB_{1} = GM_{C} \times GC_{1}$ as desired. 