Question

33. Prove that the circumcenter, O, the centroid, G, and orthocenter, H, lie on a common...

33. Prove that the circumcenter, O, the centroid, G, and orthocenter, H, lie on a common line, known as the Euler line of the triangle. (Hint:
One way to approach this proof is to construct the line containing O and G. Then find a point X on OG such that G is between O and X, and 2|OG| = |OX|. Show that this point X is on all three altitudes, and hence X is the orthocenter G.)

Homework Answers

Answer #1

Let the centroid be (G), the orthocenter (H) and the circumcenter (C).

Step1:- Let X be the midpoint of EF. Construct the median DX. Since G is the centroid, G is on DX by the definition of centroid. Also, construct the altitude DM. Since H is the orthocenter, H is on DM by the definition of orthocenter. Therefore, DM meets EF at a right angle.

Step2:- Construct a line through points C and G so that it intersects DM. Note, it appears that line CG intersects DM at the point H. However, this has not been proven yet. So, label the point of intersection H'.

Step3:- Construct CX. Since C is the circumcenter, CX meets EF at a right angle by the definition of circumcenter.

Step4:- Since DM and CX both meet EF at right angles, DM is parallel to CX. Then, the median DX is a transversal that cuts DM and CX.

Step5:-

Therefore, ∠GDH’ = ∠GXC, since they are alternate interior angles. Also, ∠DGH’ = ∠XGC, because they are vertically opposite angles.

Step6:- Since G is the centroid, DG = (2/3)DX and GX = (1/3)DX. Therefore, DG = 2GX.

Step7:-
Since ∠GDH’ = ∠GXC and ∠DGH’ = ∠XGC, ΔDGH’ ~ ΔXGC, by A-A similarity criterion. Therefore, GH’ = 2GC; since DG = 2GX.

Step8:- Thus, H' is located at the intersection of DM and GC and is in fact on the line GC such that GH' = 2GC. Thus, G, H', and C are collinear. Similarly, H' can be proved to be located on the altitudes constructed from vertices E and F so that GH' = 2GC. Therefore, H' lies on all three altitudes.

Step9:- Thus, H' is the orthocenter because it lies on all three altitudes. Yet, by the given hypothesis, H is the orthocenter. Thus, H' = H.

Step10:- Therefore, G, H, and C are collinear.


The line that connects the centroid (G), the orthocenter (H), and the circumcenter (C) is called the Euler Line.

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