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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 853.
Let FG=x, and FG meet AD at H. ThenGC=2x, AH=BG=5−2x, FH=5−x, AF=5. By Pythagoras theorem, AH^2 + FH^2 = AF^2(5-2x)^2 + (5−x)^2 = 5^2x^2 − 6x + 5 = 0(x − 1)(x − 5) = 0Since x<5, thus x=1. i.e. FG=1.
Let CF=x, FG=y and CE meet AD at H.Then ,2EC=CH and AH=5. By Pythagoras theorem of the triangle HCD , 2EC=5sqrt(5)CB^2=CF.CH , 25=x. 2ECTriangles FGC,HCD are similar.So, x/CH=y/5 ,x=CH.y/5Therefore 25=(2EC) . CH.y/5. =y.(2EC)^2/5 .So ,y=1Image: http://img802.imageshack.us/img802/7513/geogebra7.png
Reflect D along AB to have D', such that D'AD is the diameter of the circle. By pyth. thm. on CDD', we have CD' = 5*sqrt(5)By power of circle, CB*CB = CF*CP => CF = sqrt5.Since triangle(CGF)~triangle(CBE), CG:GF = CB:CE = 2:1, so again by pyth. thm. on triangle(CGF), FG = 1
FC=sqrt(5)*FG by Pythagoras. EC meets AD at P such that PD=5*2 and is a diameter of circle A because triangle PDC is similar to CBE which has a leg to leg ratio of 2. Because PD is diameter, <PFD=90. Combined with <GCF=<CDF, we know that triangle GCF similar to CDF so GF/(sqrt(5)*GF)=(sqrt(5)*GF)/5, making GF=1.