Eight Point Conic
The following problem was posed by myself and Michael Fox in the Problem Corner of the Mathematical Gazette, Nov 2006: The diagonals of a cyclic quadrilateral ABCD meet at P. The feet of the perpendiculars from P to AB, BC, CD, DA are E, F, G, H. The lines EP, FP, GP, HP meet the opposite sides of ABCD at K, L, M, N. Prove that E, F, G, H, K, L, M, N lie on a conic.
Drag A, B, C or D - also check when ABCD becomes a crossed quadrilateral.
Can you explain why (prove) the above result is true? Can you generalise further?
(Go to Further Generalizations)
Created by Michael de Villiers, 9 March 2008 with Cinderella