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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Obviously H is the orthocenter of ΔABC, and circle O is its nine-point circle. Let P be the circimcenter of ΔABC, then BH=2×PF. Now H,O,P are the Euler Line with HO=OP. Also clearly, H,G,F are the median of ΔAHC, with HG=GF. Thus, PF=2×OG, and hence BH=2×PF=4×OG.
http://s25.postimg.org/gymbfhbjz/pro_1054.pngNote that circumcircle of triangle MNF is 9 points circle of triangle ABCSince N and F are midpoints of HC and AC => NF=1/2 AH=MHAnd NF//MH => HNFM is a parallelogramSince G is the midpoint of MN => H, G, F are collinear and GF=GHSince ∠(LKF)= 90 degree => FOL is a diameter of circle O => GO= ½ GL= ¼ HB
Let BH intersect circle O (9-pt circle of ΔABC) at L.First observe Δs LMN, FMN have same circumcircle.L bisects BH. So LM // BA , so perp to HN. already BH perp MN.So H is the orthocenter of ΔLMN.Hence BH = 2 LH = 2(2 OG) = 4 OG..