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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 795.
Let KL is perpendicular to AB. Then, FBGE, CBDE and KBHF are all subscribed-able quadrilaterals. So, BH is perpendicular to CD and from this MH is perpendicular to FG. You can see this here: http://mtz256.files.wordpress.com/2012/08/7953.jpg
Proof(I will post a link with a picture):Connect CB,BD.By a property of tangent lines:∠ECD=∠CBA,∠EDC=∠ABD∵180-∠CED=∠ECD+∠EDC∴180-∠CED=∠CBA+∠ABD=∠CBD∴C,B,D,E are concyclic points∵∠EFB=∠EGB=90°∴E,F,B,G are also concyclic pointsConnect BEBy properties of concyclic figures:∠BED=∠DCB, ∠BEG=∠BFG∴∠DCB=∠BFG∴C,F,H,B are concyclic pointsConnect BH∴∠CHB=∠CFB=90°To prove that MH⊥FG, we just need to prove that ∠FHC=∠MHBBy properties of concyclic figures:∠FHC=∠FBC,∠HBF=∠FCH=∠CBA∴∠HBA=∠CBF=∠FHC∵△AHB is a right triangle and M is that midpoint of AB.∴∠MHB=∠HBA=∠FHC （Done）∴∠FHC+∠CHM=∠MHB+∠CHM=90°
Picture for the proof above:http://i1237.photobucket.com/albums/ff480/Evan_Liang/795.png
B, D, E, C are concyclic (*)So ∠BDC =∠BECB, G, E, F are concyclicSo ∠BEF = ∠BGFFollows ∠BDH = ∠BDC = ∠BEC = ∠BEF = ∠BGF= ∠BGHSo B, G, D, H are concyclic; also BGD = 90⁰Hence ∠BHD = 90⁰(*)∠CBD =∠ CBA + ∠DBA = ∠ECD +∠ EDC = 180⁰ - ∠ CED