Concurrence of Ten Nine-Point Circles
What Might This Be About?
Given point $P$ in the plane of $\Delta ABC,$ let $H$ be the orthocenter of the triangles; consider three quadrilaterals $ABPH,$ $BCPH,$ and $ACPH.$ Each quadrilateral defines four triangles whose nine-point circles concur at a point, known as the Euler-Poncelet points of the quadrilateral.
Prove that the three Euler-Poncelet points coincide.
Solution is wanting; meanwhile you may wonder why the caption says "10 circles" instead of $4\times 3=12.$
Also, it has been observed that, similar to Griffiths' theorem, that point is the same for all points $P$ on a straight line through the circumcenter of $\Delta ABC.$
The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.