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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.This entry contributed by Ajit Athle.Click the figure below to see the problem 758 details.
Let vec(BA)=a, vec(BC)=b. By Menelaus' theorem in triangle BEC and line AFD, EF/FC * CD/DB * BA/AE = EF/FC * 2/1 * 3/1 = 1so EF/FC = 1/6. vec(BE) = 2/3 avec(BF) = [6 vec(BE) + vec(BC)]/7 = (4a+b)/7vec(CE) = vec(BE) - vec(BC) = (2a-3b)/3Now let |a|=|b|=k, thena.a = k^2b.b = k^2a.b = k*k*cos(60) = k^2 / 2Therefore, vec(BF).vec(CE) = (4a+b).(2a-3b)/9= [8 a.a - 10 a.b - 3 b.b] / 9= [8 - 5 - 3]k^2 / 9= 0Hence, BF perp CE.
Haven't found the synthetic solution yet but here's my analytical attempt: WLOG assume the midpoint of AC as the origin (0,0) and let A:(-1.0) & B:(1.0) hence C:(0,√3). We can now determine D to be (1/3,2/√3) & E:(-2/3,1/√3). Slope CE=-√3 /5 and eqn CE: 5y =-√3x + √3. Similarly, slope AD = √3/2 and eqn. AD: 2y =√3x + √3. Solve the two eqns. simultaneously to obtain F as (-3/7,2√3/7). Therefore, slope BF = 5/√3 and finally (slope BF * slope CE)= -1. Hence BF _l_ CE.Will keep looking for a synthetic solution.
∆ABD≡∆CAE∴ ∠AEC = ∠ADB and so B,E,F,D are concyclic∴ ∠AFE = 60°Let M be the midpoint of BE∆BMD is equilateralSince ME = MB = MD, ∆EDB is right angled at DB,D,E,F being concyclic, we have ∠BFE = ∠BDE = 90°
Tr.s CAE & ABD are congruent. So Tr.s AEF & ABD are similar since Tr.s AEF & CAE are similar. Hence BEFD is con cyclic. Now consider Tr. BDE where BE = 2. BD & the included < is 60. So Tr. BDE is a 30-60-90 Tr. and the result follows.Sumith PeirisMoratuwaSri Lanka