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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 716 details.
Some clarifications and minor corrections of previous solution:http://img849.imageshack.us/img849/782/problem716.pngDraw diameter FK of circle O’ . Connect KA and KB KA perpen to FA , KB perpen to FBFrom G &H draw GA’//KA and HB’//KB ( see picture) . GA’ cut FK at NNote that GA’ perpen to FA , HB’ perpen to FB1. Triangle FAG congruence to FBH (case AA)Let FG/FA=FH/FB= k and m(HFB)=m(GFA)=xSo FA’/FA= FG/FA . cos(x)= k.cos(x)= FN/FK And FB’/FB=FH/FB. cos(x)= k.cos(x) =FA’/FA=FN/FKSo GA’ and HB’ will intersect at point N on FK (properties of // lines and congruence triangles) 2. In triangle GHN, HA’ and GB’ are altitudes and F will be the orthocenter of triangle GHN so O’F perpen To GHWith the same reason as part 1, OC will perpen to ED and L will be the orthocenter of triangle C’MF’And ML will perpen to C’F’Peter Tran
Join BA and extend it.Draw tangent at F to circle (O’) to meet BA extended at X.Let tangent XF be extended to meet C’F’ at Y.∠YFB = ∠FAB (angle in the alternate segment).= ∠HAB = ∠HGB (angles in the same segment).∴ XY ∥ MH. In turn this implies F’O’F ⊥ MH.By a similar argument C’OC ⊥ MD.∴ L is the orthocenter of ∆ MC’F’ and Hence ML ⊥ C’F’
Very good solution Pravin !Peter Tran
Let CC’ meet circle O at P and DE at Q. Let F’F meet HG at R.< EDA = < CBA = < CPA, hence APDQ is concyclic and so < PQF’ = < PAC = 90.Similarly we can show that < C’RF = 90Hence L is the orthocenter of Triangle MC’F’ and so ML is perpendicular to C’F’Sumith PeirisMoratuwaSri Lanka