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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 785.
Let K be the mid-point of AC, such that BK is the altitude. Note that triangles ADE and EFC are both equilateral, and also DBFE is a parallelogram, soAE = DE = BFEC = EF = DBThusa + d + f= d + EC + AE + f= d + DB + BF + f= GB + BH= 2*MK + 2*KN= 2*(MK + KN)= 2x
Let AE = b, then AD = b, EC = a-b, FC = a-bThen AM = ADcos60 = (b-d)/2NC = CFcos60 = (a-b-f)/2by considering the length of AC = AM + MN + NC, a = (b-d)/2 + x + (a-b-f)/2 x = (a+d+f)/2q.e.d.
Triangles ADE and CFE are equilateralSo DE + EF = AE + EC = aDraw DJ, FK each perpendicular to ACx = ME + EN = (MJ + JE)+(EK + KN)= [(GD + DE) + (EF + FH)]cos 60°= [GD + (DE + EF) + FH]/2= (d + a + f)/2