Friday, May 11, 2012

Problem 752: Square, Circle, Tangent, Chord, Angle

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

Click the figure below to see the problem 752 details.

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

2. 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

3. Join CE, CF
Rt ∆BEC ≡ Rt ∆DFC (∵ BC = CD, CE = CF)
∴∠CDF = ∠CBE = 25°
∠ECD = ∠BCD - ∠BCE = 90° - ∠ECB = 25°
∠DCF = 90° - ∠CDF = 90°- 25° = 65°
CE = CF implies ∠CFE = ∠CEF = 45°
∴ ∠EFD = 45° (∵ CF ⊥ DF)
Hence x = ∠CDF + ∠EFD = 25° + 45° = 70°

4. 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.

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

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

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

Sumith Peiris
Moratuwa
Sri Lanka