Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 752 details.

## Friday, May 11, 2012

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http://img526.imageshack.us/img526/9787/problem752.png

ReplyDeleteConnect CE and CF

Note that CE perpendicular to BE and CF perpendicular to FD

∠CDF=∠ECG=25 and ∠DCF=65=> ECF is a right isosceles triangle

In triangle ECG, external angle x= ∠CEG+∠ECG= 70

Triangle DCF and ACE are congruent triangles.

ReplyDeleteHence 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, CF

ReplyDeleteRt ∆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,

ReplyDeleteSo ECF is a right isoceles triangle.

x = 45 deg. + angle ECG = 70 deg.

http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27270&p=133540

ReplyDeleteTr.s BCE and DCF are congruent so < CDF = 25 =< ECF. Also< DCF = 65

ReplyDeleteHence Tr. ECF is an isoceles right Tr. and so x=70

Sumith Peiris

Moratuwa

Sri Lanka