Solution

The solution is by Michael Lambrou posted at the Hyacinthos forum.

The same could be said a little differently. Consider a point P' in 3D such that PP' is perpendicular to the plane of ABC. Let points X', Y', and Z' be symmetries of P' in the midpoints of CB, BA, and AC, respectively. Then points A, B, C, P', X', Y', Z' serve as vertices of a parallelepiped. The required concurrence of AX', BY' and CZ' is obvious. Now observe that the original configuration is the orthogonal projection of that parallelepiped onto the plane. The conclusion follows since projection preserves concurrence.

Note that our approach makes it also obvious that triangle ABC and XYZ are in fact equal.

There are further examples where stepping out of a 2 dimensional plane into 3 dimensions makes a problem at hand easily solvable:

2D problems that benefit from a 3D outlook