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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 752 details.
http://img526.imageshack.us/img526/9787/problem752.pngConnect CE and CFNote that CE perpendicular to BE and CF perpendicular to FD∠CDF=∠ECG=25 and ∠DCF=65=> ECF is a right isosceles triangleIn triangle ECG, external angle x= ∠CEG+∠ECG= 70
Triangle DCF and ACE are congruent triangles.Hence Angle CDF = 25.Let x be angle DFG, then angle ECF = 2x -> angle ECG = 2x - 55. Then angle BCD = 90 = 55 + (2x -55 ) Therefore x = 45. By external angle, CGF = 45 + 25 = 70
Join CE, CFRt ∆BEC ≡ Rt ∆DFC (∵ BC = CD, CE = CF)∴∠CDF = ∠CBE = 25°∠ECD = ∠BCD - ∠BCE = 90° - ∠ECB = 25°∠DCF = 90° - ∠CDF = 90°- 25° = 65° Adding: ∠ECF = 90°CE = CF implies ∠CFE = ∠CEF = 45°∴ ∠EFD = 45° (∵ CF ⊥ DF)Hence x = ∠CDF + ∠EFD = 25° + 45° = 70°
Right triangle CFD is formed by rotating right triangle CEB about C for 90 deg, So ECF is a right isoceles triangle.x = 45 deg. + angle ECG = 70 deg.
Tr.s BCE and DCF are congruent so < CDF = 25 =< ECF. Also< DCF = 65Hence Tr. ECF is an isoceles right Tr. and so x=70Sumith PeirisMoratuwaSri Lanka