সম্প্রসারণ: ত্রিভুজ কেন্দ্র &O,H,N,G& সমরেখ ও অয়লার রেখায় বিদ্যমান

The Euler line is a very special line that exists uniquely in each triangle (except equilateral ones).

It's commonly defined to be the line $OH$.

Recall from our investigation of the nine point circle that N is the midpoint of OH. (You can read it in the nine-point circle topic). So obviously, N, O, H lie on the Euler line. (O = circumcenter, H = orthrocenter)

So the title tells us that G is on the Euler line.

Let's try to prove this.

We can do it with a simple application of vectors:

Lowercase letters denote the vectors of the points (a is vector A for example)

Since the origin is anywhere and we are lazy people, we'll assume the origin is at H for convenience.

So then $g = \frac{a+b+c}{3}$ by a well-known property of the centroid. All we need is to find o

Let's say D is the midpoint of BC, E is the midpoint of AC, F is the midpoint of AB. Obviously, $d = \frac{b+c}{2}, e= \frac{a+c}{2}, f = \frac{a+b}{2}$.

We can do some calculations (I don't want to too deeply into vectors, so I won't post them here) to find that $o = \frac{a+b+c}{2}$.

Now... we have $h = 0(a+b+c), g = \frac{1}{3} (a+b+c), o = 0.5 (a+b+c)$... This means that $O,G,H$ are collinear! In fact, since O and H are on the Euler line, this means that $G$ is on the Euler line.

Let's analyze the Euler line for special triangles.

1. Equilateral

Since H,O,N,G are all the same point in an equilateral triangle, the Euler line doesn't even exist!

1. Isosceles but not equilateral

We can easily find (try drawing it) that the Euler line is the symmetric axis of the isosceles triangle (in other words, if AB=AC, the Euler line is the perpendicular bisector of BC)

1. Right triangle

You can draw this one too. The Euler line is the median from the right angle to the hypotenuse (of course, the median can be extended)

Another property of the Euler line is that if it is parallel to side BC of a triangle, then $\tan B \tan C = 3$.

In every triangle, there is a line that contains the circumcenter $O$, orthocenter $H$, nine-point center $N$, and the centroid $G$.

First, start by proving that the circumcenter, centroid, and the nine-point center are collinear. (Hint: Look for a homothety)

Proof:

Notice that there exists a homothety mapping the triangle to its medial triangle centered at $G$. Thus the circumcenter is mapped to the nine-point center (which is the circumcenter of the medial triangle), and $O$, $G$, and $N$ are collinear. $\Box$

Now we will prove that the orthocenter, the circumcenter, and the centroid are collinear. (Hint: This is very similar to the previous)

Proof: Draw lines $d$, $e$, $f$ through $A$, $B$, $C$ parallel to $BC$, $AC$, $AB$, respectively. Let $e$ and $f$ intersect at $D$, $d$ and $f$ at $E$, and $d$ and $e$ at $F$.

Since $DBAC$ and $FBCA$ are parallelograms, we have that $DB = AC = BF$. Thus, $B$ is the midpoint of $DF$. Similarly, $A$ is the midpoint of $EF$ and $C$ is the midpoint of $DE$. Thus $ABC$ is the medial triangle of $DEF$, and thus there exists a homothety that maps the triangle to its medial triangle centered at $G$. Then $H$ is mapped to $O$, and $H$, $G$, $O$ are collinear. $\Box$

We combine the two facts and we deduce that $O$, $H$, $G$, and $N$ are collinear.