Euler line

In any triangle $\triangle ABC$ , the Euler line is a line which passes through the orthocenter $H$ , centroid $G$ , circumcenter $O$ , nine-point center $N$ and de Longchamps point $L$ . It is named after Leonhard Euler. Its existence is a non-trivial fact of Euclidean geometry. Certain fixed orders and distance ratios hold among these points. In particular, $\overline{OGNH}$ and $OG:GN:NH = 2:1:3$

Euler line is the central line $L_{647}$ .

Given the orthic triangle $\triangle H_AH_BH_C$ of $\triangle ABC$ , the Euler lines of $\triangle AH_BH_C$ , $\triangle BH_CH_A$ , and $\triangle CH_AH_B$ concur at $N$ , the nine-point circle of $\triangle ABC$ .

Proof Centroid Lies on Euler Line

This proof utilizes the concept of spiral similarity, which in this case is a rotation followed homothety. Consider the medial triangle $\triangle O_AO_BO_C$ . It is similar to $\triangle ABC$ . Specifically, a rotation of $180^\circ$ about the midpoint of $O_BO_C$ followed by a homothety with scale factor $2$ centered at $A$ brings $\triangle ABC \to \triangle O_AO_BO_C$ . Let us examine what else this transformation, which we denote as $\mathcal{S}$ , will do.

It turns out $O$ is the orthocenter, and $G$ is the centroid of $\triangle O_AO_BO_C$ . Thus, $\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}$ . As a homothety preserves angles, it follows that $\measuredangle O_AOG = \measuredangle AHG$ . Finally, as $\overline{AH} || \overline{O_AO}$ it follows that $\triangle AHG = \triangle O_AOG$ Thus, $O, G, H$ are collinear, and $\frac{OG}{HG} = \frac{1}{2}$ .

Another Proof

Let $M$ be the midpoint of $BC$ . Extend $CG$ past $G$ to point $H'$ such that $CG = \frac{1}{2} GH$ . We will show $H'$ is the orthocenter. Consider triangles $MGO$ and $AGH'$ . Since $\frac{MG}{GA}=\frac{H'G}{GC} = \frac{1}{2}$ , and they both share a vertical angle, they are similar by SAS similarity. Thus, $AH' \parallel OM \perp BC$ , so $H'$ lies on the $A$ altitude of $\triangle ABC$ . We can analogously show that $H'$ also lies on the $B$ and $C$ altitudes, so $H'$ is the orthocenter.

Proof Nine-Point Center Lies on Euler Line

Assuming that the nine point circle exists and that $N$ is the center, note that a homothety centered at $H$ with factor $2$ brings the Euler points $\{E_A, E_B, E_C\}$ onto the circumcircle of $\triangle ABC$ . Thus, it brings the nine-point circle to the circumcircle. Additionally, $N$ should be sent to $O$ , thus $N \in \overline{HO}$ and $\frac{HN}{ON} = 1$ .

Analytic Proof of Existence

Let the circumcenter be represented by the vector $O = (0, 0)$ , and let vectors $A,B,C$ correspond to the vertices of the triangle. It is well known the that the orthocenter is $H = A+B+C$ and the centroid is $G = \frac{A+B+C}{3}$ . Thus, $O, G, H$ are collinear and $\frac{OG}{HG} = \frac{1}{2}$

Euler line for a triangle with an angle of 120 $^\circ.$

Let the $\angle C$ in triangle $ABC$ be $120^\circ.$ Then the Euler line of the $\triangle ABC$ is parallel to the bisector of $\angle C.$

Proof

Let $\omega$ be circumcircle of $\triangle ABC.$

Let $O$ be circumcenter of $\triangle ABC.$

Let $\omega'$ be the circle symmetric to $\omega$ with respect to $AB.$

Let $E$ be the point symmetric to $O$ with respect to $AB.$

The $\angle C = 120^\circ \implies O$ lies on $\omega', E$ lies on $\omega.$

$EO$ is the radius of $\omega$ and $\omega' \implies$ translation vector $\omega'$ to $\omega$ is $\vec {EO}.$

Let $H'$ be the point symmetric to $H$ with respect to $AB.$ Well known that $H'$ lies on $\omega.$ Therefore point $H$ lies on $\omega'.$

Point $C$ lies on $\omega, CH || OE \implies CH = OE.$

Let $CD$ be the bisector of $\angle C \implies E,O,D$ are concurrent. $OD = HC, OD||HC \implies CD || HO \implies$

Euler line $HO$ of the $\triangle ABC$ is parallel to the bisector $CD$ of $\angle C$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

Concurrent Euler lines and Fermat points

Consider a triangle $ABC$ with Fermat–Torricelli points $F$ and $F'.$ The Euler lines of the $10$ triangles with vertices chosen from $A, B, C, F,$ and $F'$ are concurrent at the centroid $G$ of triangle $ABC.$ We denote centroids by $g$ , circumcenters by $o.$ We use red color for points and lines of triangles $F**,$ green color for triangles $F'**,$ and blue color for triangles $FF'*.$

Case 1

Let $F$ be the first Fermat point of $\triangle ABC$ maximum angle of which smaller then $120^\circ.$ Then the centroid of triangle $ABC$ lies on Euler line of the $\triangle ABF.$ The pairwise angles between these Euler lines are equal $60^\circ.$

Proof

Let $G', O,$ and $\omega$ be centroid, circumcenter, and circumcircle of $\triangle ABF,$ respectevely. Let $\triangle ABD$ be external for $\triangle ABC$ equilateral triangle $\implies F = CD \cap \omega.$ $\angle AFB = 120^\circ \implies AFBD$ is cyclic.

Point $O$ is centroid of $\triangle ABD \implies \vec O = \frac {\vec A + \vec B + \vec D}{3}.$ $\vec G' = \frac {A + B + F}{3}, G = \frac {A + B + C}{3} \implies$ $\vec {OG} = \frac {\vec C – \vec D}{3} = \frac {\vec DC}{3}, \vec {G'G} = \frac {\vec C – \vec F}{3} = \frac {\vec FC}{3} \implies$ $OG||G'G \implies$ Points $O, G',$ and $G$ are colinear, so point $G$ lies on Euler line $OG'$ of $\triangle ABF.$

$\vec {GG_0} = \frac {A – F}{3} \implies GG_0||AF, \vec {GG_1} = \frac {B – F}{3} \implies GG_1||BF.$

Case 2

Let $F$ be the first Fermat point of $\triangle ABC, \angle BAC > 120^\circ.$ Then the centroid $G$ of triangle $ABC$ lies on Euler line of the $\triangle ABF.$

Proof

Let $\triangle ABD$ be external for $\triangle ABC$ equilateral triangle, $\omega$ be circumcircle of $\triangle ABD \implies F = CD \cap \omega.$ $\angle ABD = 60^\circ, \angle AFD = 120^\circ \implies ABDF$ is cyclic.

Point $O$ is centroid of $\triangle ABD \implies$ $\vec {OG} = \frac {\vec DC}{3}, \vec {G'G} = \frac {\vec FC}{3} \implies OG||G'G \implies$

Points $O, G',$ and $G$ are colinear, so point $G$ lies on Euler line $OG'$ of $\triangle ABF$ as desired.

Case 3

Let $F'$ be the second Fermat point of $\triangle ABC.$ Then the centroid $G$ of triangle $ABC$ lies on Euler lines of the triangles $\triangle ABF',\triangle ACF',$ and $\triangle BCF'.$

Proof

Let $\triangle ABD$ be internal for $\triangle ABC$ equilateral triangle, $\omega$ be circumcircle of $\triangle ABD \implies F' = CD \cap \omega.$

Let $O_1, O_0,$ and $O'$ be circumcenters of the triangles $\triangle ABF',\triangle ACF',$ and $\triangle BCF'.$ Point $O_1$ is centroid of the $\triangle ABD \implies GO_1G_1$ is the Euler line of the $\triangle ABF'$ parallel to $CD.$

$O_1 O_0$ is bisector of $BF', O'O_1$ is bisector of $AF', O'O_0$ is bisector of $CF' \implies \triangle O'O_1O_0$ is regular triangle.

$\triangle O'O_0 O_1$ is the inner Napoleon triangle of the $\triangle ABC \implies G$ is centroid of this regular triangle. $\angle GO_1O_0 = 30^\circ, O_1O_0 \perp F'B, \angle AF'B = 120^\circ \implies GO_0||F'A.$

$\vec {GG_0} = \frac {\vec {F'A}}{3} \implies GG_0||F'A \implies$ points $O_0,G,$ and $G_0$ are collinear as desired.

Similarly, points $O',G,$ and $G'$ are collinear.

Case 4

Let $F$ and $F'$ be the Fermat points of $\triangle ABC.$ Then the centroid of $\triangle ABC$ point $G$ lies on Euler line $OG' (O$ is circumcenter, $G'$ is centroid) of the $\triangle AFF'.$

Proof

Step 1. We find line $F'D$ which is parallel to $GG'.$

Let $M$ be midpoint of $BC.$ Let $M'$ be the midpoint of $FF'.$

Let $D$ be point symmetrical to $F$ with respect to $M.$

$MM'||DF'$ as midline of $\triangle FF'D.$ $\vec {G'G} = \frac {\vec A + \vec B + \vec C}{3} – \frac {\vec A + \vec F + \vec F'}{3} = \frac {2}{3} \cdot (\frac {\vec B + \vec C}{2} – \frac {\vec F + \vec F'}{3})$ $\vec {G'G} = \frac {2}{3} (\vec M – \vec M') = \frac {2}{3} \vec {M'M} \implies M'M||F'D||G'G.$

Step 2. We prove that line $F'D$ is parallel to $OG.$

Let $\triangle xyz$ be the inner Napoleon triangle. Let $\triangle XYZ$ be the outer Napoleon triangle. These triangles are regular centered at $G.$

Points $O, z,$ and $x$ are collinear (they lies on bisector $AF').$

Points $O, Z,$ and $Y$ are collinear (they lies on bisector $AF).$

Points $M, X,$ and $y$ are collinear (they lies on bisector $BC).$ $E = YZ \cap BF, E' = Zx \cap BC.$ $BF \perp XZ \implies \angle BEZ = 30^\circ.$ $BC \perp Xy,$ angle between $Zx$ and $Xy$ is $60^\circ \implies \angle BE'Z = 30^\circ.$

$\angle AF'B = 120^\circ, \overset{\Large\frown} {AZ} = 60^\circ \implies$

Points $A, Z, F', B, E',$ and $E$ are concyclic $\implies \angle OZx = \angle CBF.$ $FM = MD, BM = MC \implies \angle CBF = \angle BCD.$ Points $C, D, X, B,$ and $F'$ are concyclic $\implies \angle BCD = \angle BF'D.$

$\angle GZO = \angle GxO = 30^\circ \implies$ points $Z, O, G,$ and $x$ are concyclic

$\implies \angle GOx = \angle OZx – 30^\circ = \angle AF'B – 30^\circ.$ $\angle AF'B = 120^\circ, Ox \perp AF' \implies OG||F'D.$ Therefore $OG||G'G \implies O, G',$ and $G$ are collinear or point $G$ lies on Euler line $OG'.$