From a point P in the plane of ABC drop perpendiculars to the side lines of the triangle. Pass a circle through the feet of the three perpendiculars. In general, the circle will cross the side lines at another three points. At those points erect perpendiculars to the side lines. The latter three lines always concur at a point.
The proof is simple. The point of concurrence is the symmetric image of P in the center of the circle.