Triangles with common base

An interesting problem proposed by Steve Gray

Drag the point $D$.
Press “r” to reset the diagram to its initial state.
Proposition: $A'B'C'$ is equilateral and its centroid $M$ is fixed.

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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