Show that the centers of the squares on the outside of the four sides of a rhombus form a square.
Solution
Take the rhombus as ABCD and the centers as PQRS as shown. Let the center of the rhombus be O. By symmetry about AC, we have OP = OS. Now ∠AOD = ∠ ASD = 90o, so AODS is cyclic. Hence ∠AOS = ∠ADS = 45o. Similarly for all the other angles. So POR and QOS are straight and perpendicular. Hence PQRS is a square.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03