# Existence of the Euler Line: An Elementary Proof

In any triangle, the orthocenter, the centroid, and the circumcenter are collinear on the Euler line. In addition GH = 2GO.

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Discussion The proof of the existence of the Euler line illustrated by the applet is due to Vladimir Nikolin, an elementary school teacher from Serbia.

The key to the proof is the observation that the antipodes of the vertices in the circumcircle along with the orthocenter and two vertices form parallelograms whose diagonals intersect at the midpoints of the sides of the triangle. The applet highlights those parallelograms - one at a time - depending on which of the vertices of the triangle is being selected or dragged. (The "hint" button needs to be checked for the parallelograms to be shown.)

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

For the definiteness sake, let's focus on vertex C and its antipode F, so that CF is a diameter of the circumcircle.

Since CF is a diameter, ∠CBF = 90° (as an inscribed angle subtended by a diameter), implying BF⊥BC. Also, AH being an altitude in ΔABC, is perpendicular to BC: AH⊥BC. Therefore, AH||BF.

Similarly, BH is an altitude (BH⊥AC) while ∠CAF = 90°. Therefore AF||BH.

The quadrilateral AFBH is a parallelogram. The diagonals AB and FH cross at their common midpoint which means that the midpoint M c of AB lies on FH and, in addition, HM c = M cF.

It follows that CM c serves as a median in both triangles ABC and HFC. In ΔABC, CM c is divided by the centroid G in the ratio 2:1. In ΔHFC, O being the midpoint of side CF, HO is another median which meets CM c at the centroid of ΔHFC. The latter divides CM c in the ratio 2:1, meaning that it coincides with G and implying that the three points H, G, O are collinear and that GH = 2GO. • 