Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 611.
Analytic Proof:Let D = (o,o), A = (a,0), C = (c,0), B = (0,b) The equations of AB, DEM are respectivelybx + ay = ab and ax-by = 0Solving, we haveE = [ab²/(a² + b²) , a²b/(a² + b²)]Similarly,F = [b²c/ (b² + c²), bc²/ (b² + c²)]It can be verified that the equation of EF is b (a + c)x – (b² - ac)y = abcHence G = [0, abc/ (ac - b²)] and so the equation of CG is abx + (ac - b²)y =abcSolving it with the equation ax-by = 0 of DEwe obtain (eliminating x)y-coord of M = b = y-coord of BM and B are at the same height over ACHence BM is parallel to ACSimilarly BH is parallel to ACThus M, B, H are collinear
http://img840.imageshack.us/img840/7563/problem611.pngDraw circle diameter BD ( see attached sketch)From B draw a line parallel to AC . This line cut DE and DF at M and HAH cut BD at G and MC cut BD at G’ .By the properties of right triangles we have BH.DC=BD^2= BM.ADSo BM/DC=BH/AD∆MBG similar to ∆ CDG => BG’/DG’= BM/DC∆HBG similar to ∆ADG => BG/DG= BH/ADFrom above expressions we have BG’/DG’= BG/DG => G coincide to G’M, B and H are collinear and BH // AC