Triangles with common base

An interesting problem proposed by Steve Gray

Sorry, no pictures. Explanation here.

The construction

This problem was proposed by Steve Gray in the geometry.puzzles newsgroup (see the original message) on July 26, 2002. Scroll to the bottom of that page for a link to the solution.

Consider an equilateral triangle $ABC$, a line segment $PQ$, and an arbitrary point $D$, as seen in the diagram above. On the segment $PQ$ construct three triangles $PC'Q$, $PA'Q$, $PB'Q$, similar to the triangles $ADB$, $BDC$, $CDA$, respectively.

Proposition 1: The triangle $A'B'C'$ is equilateral.

Proposition 2: The centroid of $A'B'C'$ is independent of $D$.

Steve adds:

Now generalize this for a regular $n$-gon. The new points form a regular $n$-gon whose centroid is independent of $D$. This problem is original so far as I know. I am interested in the simplest synthetic solution; no algebra, please.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet July 26, 2002.
Cosmetic revisions on June 17, 2010.

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