Euler-Poncelet Point

What Might This Be About?

10 March 2014, Created with GeoGebra


Given quadrilateral $ABCD,$ denote the nine-point circles of triangles $ABD,ABC,BCD,ACD$  just as $1,2,3,4,$ respectively.

Euler-Poncelet Point - problem

The four circles meet at a point.

The point is variably known as the Euler, Euler-Poncelet, and Poncelet point.


Let points $I,$ $J,$ $K,$ $L,$ $M,$ $N$ be the midpoints of $AB$, $BC$, $CD$, $DA$, $AC$, $BD,$ respectively. Let $P$ be the second intersection of circles $3$ and $4.$

Euler-Poncelet Point - solution

Being the midlines in triangles $ACD$ and $BCD$ parallel to the base $CD,$ $LM\parallel JN.$ Similarly $IM\parallel KN.$ This implies $\angle LMI=\angle JNK.$ As inscribed angles subtended by the same chord, $\angle JNK=\angle JPK.$ By transitivity, $\angle LMI=\angle JPK.$

Due to Varignon's observation, $LK\parallel IJ.$

Now focus on circles $2$ and $4.$ By one of Reim's theorems, since $\angle LMI=\angle JPK,$ points $M,P,I,J$ are concyclic. But $M,P,I$ lie on circle $2,$ implying that so does $P.$ It follows that circles $2,3,4$ concur at point $P.$

Similarly, circles $1,3,4$ meet at $P;$ therefore, $P$ is common to all four circles.


The existence of the Euler-Poncelet point has been the subject of an earlier page, with a reference to a proof in complex variables by J. L. Coolidge. The above proof is due to Jean-Louis Ayme and is available on the web in an article of his, where one (weaker) version of Reim's theorem has been proved but a stronger version used.

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