If a cyclic quadrilateral ( = with vertices lying on a common circle) has diagonals which are perpendicular, then the perpendicular to a side from the point of intersection of the diagonals will bisect the opposite side (AF = FD).
The three blue points always lie on a straight line. The blue points are the closest points to the moving red point on the lines. In other words the blue points are the projections of the moving red point to the lines.
Clever visual proof by Mike Hirschhorn.
If you place squares on the sides of any parallelogram, their centers will always form a square.
Given 3 circles, each intersecting the other two in two points, the line segments connecting their points of intersection satisfy: ace/bdf = 1
In any triangle, the 3 points of intersection of the adjacent angle trisectors ALWAYS form an equilateral triangle (in blue), called the Morley triangle.