There’s far too little geometry—excluding topology and non-Euclidean stuff—on this blog, so let’s add a little.

Our goal is to get to the Euler line, a line that passes through a triangle’s circumcenter, centroid, and orthocenter. The line is only determined for non-equilateral triangles; the points coincide in the equilateral case. We’ll look at the three points above.

The circumcenter, centroid, and orthocenter are all “centers” of triangle. But what is a center of a triangle? Surely, it’s not a point equidistant to all points on the triangle. Our triangle would be a circle in that case.

The circumcenter of a triangle *ABC* is the center *O* of the circle *K* that triangle *ABC* is inscribed in.

The circumcenter is actually the intersection of the three perpendicular bisectors of the triangle: *FE*, *IG*, and *DH*. To see this, first suppose that triangle *ABC* has a circumscribed circle *K* with center *O*. Draw radii *AO*, *BO, *and *CO* to each of the triangles vertices. This creates three smaller triangles *AOB*, *BOC*, and *AOC*. In each of these smaller triangles, drop an altitude from *O*. For example, in triangle *AOB**, *altitude *OD* would be dropped. This splits *AOB* into two smaller triangles that are congruent by SAS, Line *OD* is perpendicular to *AB* by construction, and *AD = DB*. Hence *OD* is indeed a perpendicular bisector of side *AB*. Repeating this for other sides shows that the center of the circumscribed circle is the intersection of *ABC*‘s perpendicular bisectors.

Moreover, the intersection of any to perpendicular bisectors is equidistant from each of the triangle’s vertices. The reader can see this by considering triangle *AOC*. Perpendicular bisector *IG* splits *AOC* into triangles that are congruent by SAS. It follows that lengths *AO* and *OC* are equal. Repeat for the other sides. We then see that the intersection of the perpendicular bisectors is equidistant from the triangle’s vertices. Thus the perpendicular bisectors of a triangle uniquely determine its circumcenter.

The centroid is the intersection of a triangle’s three medians, lines drawn from a vertex that bisect the opposite side. As said in class, the centroid is the center of mass for a thin, triangular solid with uniformly distributed mass.

The reader may suspect whether the three medians of a triangle intersect. Clearly two of the medians intersect; otherwise our triangle *ABC* would be a line. But the full proof is a little tedious. The proof involves assuming that two medians *AF* and *CE* intersect and drawing a parallelogram using the midpoints of the medians. We link to some proofs: http://jwilson.coe.uga.edu/EMAT6680Fa06/Chitsonga/MEDIAN/THE%20MEDIANS%20OF%20A%20TRIANGLE.htm uses classical geometry and http://math.stackexchange.com/questions/432143/prove-analytically-the-medians-of-a-triangle-intersect-in-a-trisection-point-of uses vectors.

Interestingly, the midpoints of the sides of triangle *ABC*—the ends of the medians—cut the triangle into four congruent triangles. We will prove this in a roundabout way. Let *E* be the midpoint of *AB*. Draw a line *EF* parallel to *AC* where *F* intersects *BC*. Similarly draw *FD* parallel to *AB*. By construction, *EFDA* and *EFCD* are parallelograms. Then *AD = EF = DC*, so *D* is the midpoint of *AC*. Similarly, *F *is the midpoint of *BC*. The reader can see that the triangles are congruent by repeatedly applying SAS.

Our final center is the orthocenter, the intersection of the three altitudes of a triangle. An altitude is a segment drawn from a vertex that is perpendicular to the opposite side. As with the two previous centers, the intersection of the altitudes at a single point isn’t immediately obvious.

We show that the altitudes of triangle *ABC* intersect. Construct triangle *DEF* with triangle *ABC* inscribed in it by making sides *DF*, *FE*, and *DE* parallel respectively to *BC, **AB*, and *AC*. Draw altitude *BK* where *K *intersects *DF*. Since *AC* is parallel to *DE*, *BK* is perpendicular to *DE*. Moreover, *ADBC* and *BACE* are parallelograms, so *DB = AC = AE*. Hence *BK* is a perpendicular bisector of *DE*. We repeat the argument for the other altitudes of triangle *ABC*. Then the altitudes of *ABC* intersect because the perpendicular bisectors of *DEF *intersect.

There are a few other centers of a triangle that are either irrelevant to the Euler line or take too long to construct (i.e. I’m tired of drawing diagrams). The incenter is the center of the circle inscribed within a triangle. The incenter also turns out to be the center of a triangle’s angle bisectors. The Euler line doesn’t pass through the incenter.

The nine-point circle is the circle that passes through the feet of the altitudes (the end that isn’t the vertex) of a triangle.

Strangely, the circle also passes through the midpoints of the sides of its triangle. But that’s not all. The circle passes through the Euler points, the midpoints of the segments joining the triangle’s vertices to the triangle’s orthocenter. Thus the nine-point circle does indeed pass through nine special points of a triangle. The center of the nine-point circle lies on the Euler line.

After all this, we still haven’t proved that the circumcenter, centroid, and orthocenter lie on the same line. We won’t prove this. Here’s a video of the proof by Salman Khan: https://www.youtube.com/watch?v=t_EgAi574sM. The proof uses a few facts about the centers we haven’t discussed, but these facts aren’t too hard to show. Refer back to my four congruent triangles picture. Let *O*, *K*, and *L* respectively be the circumcenter, centroid, and orthocenter of triangle *ABC*. Then Khan proves that triangle *DOK* is similar to triangle *BLK*. This implies angles *DKO* and *CKL* are equal, which means *O, K, *and *L* lie on the same line.

Sources and cool stuff:

H.S.M. Coxeter and Samuel L. Greitzer’s *Geometry Revisited*

Paul Zeitz’s *The Art and Craft of Problem Solving* (Chapter 8 is called “Geometry for Americans”)

Wolfram on the nine-point circle: http://mathworld.wolfram.com/Nine-PointCircle.html

A fun way to play with the Euler line: http://www.mathopenref.com/eulerline.html

Khan’s Euler line video: https://www.youtube.com/watch?v=t_EgAi574sM

Wolfram on the Euler line: http://mathworld.wolfram.com/EulerLine.html

Classical median proof: http://jwilson.coe.uga.edu/EMAT6680Fa06/Chitsonga/MEDIAN/THE%20MEDIANS%20OF%20A%20TRIANGLE.htm

Vector median proof: http://math.stackexchange.com/questions/432143/prove-analytically-the-medians-of-a-triangle-intersect-in-a-trisection-point-of