14th ASU 1980 Problem 15 ABC is equilateral. A line parallel to AC meets AB at M and BC at P. D is the center of the equilateral triangle BMP. E is the midpoint of AP. Find the angles of DEC.
D = 60 o , E = 90 o Solution Thanks to Achilleas Sinefakopoulos for this Let K be the midpoint of BP and L the midpoint of AC. EL is parallel to BC, so ∠ELC = 120 o . EK is parallel to AB, so ∠EKC = 60 o , so ELCK is cyclic. But ∠DKC = ∠DLC = 90 o , so DLCK is cyclic. Hence D, K, C, L, E all lie on a circle. Hence ∠DEC = ∠DLC = 90 o , and ∠EDC = ∠EKC = 60 o . 14th ASU 1980 © John Scholes jscholes@kalva.demon.co.uk 26 June 2002