The Euler Segment: Why?

In the diagram, M is the midpoint of segment QR, B is the centroid of triangle PQR, and C is the circumcenter of triangle PQR.

X is a point constructed on the line BC, twice as far from B as C, and on the opposite side of B.

If we can show that X lies on the altitude from vertex P to side QR, then interchanging the roles of P, Q, and R, X will also lie on the altitudes from Q to RP, and from R to PQ. In other words, X must then be the orthocenter of triangle PQR.

First of all, angles CBM and XBP form a linear pair so they are equal.

Now remember that the medians of a triangle trisect each other at the centroid. Therefore, the vertex P is twice as far from the centroid B along the median PM as the midpoint M of the side QR.

Since B divides segment CX in the same proportion (by construction) as it does the median PM, triangles CBM and XBP are similar, and lines CM and PX are parallel.

By its definition, the circumcenter of a triangle is the intersection of the perpendicular bisectors of the sides. Therefore, CM is perpendicular to QR and PX, being parallel to CM, is too. In other words, X does lie on the altitude from vertex P to side QR:
CX is the Euler segment of triangle PQR.