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A few words

### 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.

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Given ΔABD and a point C on its base line BD. The triangles ABC and ACD share a side AC and a base line. In each of the latter two triangles, consider Euler's lines. I claim that if the Euler line in ΔACD is parallel to AB, then the Euler line in ΔABC is parallel to AD. Furthermore when this happens, the two lines meet on BD.

To show that this is indeed so I make a reference to another statement proven for the same triangle configuration. Draw a line through the centroid of ΔACD parallel to AB, and another one through the centroid of ΔABC parallel to AD. Those two lines intersect on BD. A similar statement holds for, say, the lines drawn through the circumcenters.

We may also change the order of actions a little. Draw a line parallel to AB through the centroid of ΔACD to its intersection with BD and join the latter point to the centroid of ΔABC. The resulting line is then parallel to AD. A similar statement is of course true for the circumcenters.

Therefore, if the line parallel to AB happens to be the Euler line of ΔACD, so that it passes through both the centroid and the circumcenter of ΔACD, then the lines joining its intersection with BD to the centroid and the circumcenter of ΔABC are both parallel to AD and hence coalesce into a single line which is the Euler line of that triangle.