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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 709.
http://img607.imageshack.us/img607/5235/problem709.pngExtend AF and BF to AD and BC( see picture)Let AD=BC=DN=CL=2bLet CD=2a, AF=FL=2cNote that triangle AHD similar to triangle EHL soAH/HL=AD/EL=HD/HE=2/3From this ratio we get HF=2c/5 and AH/HF=4Triangle BGE similar to triangle NGA so AG/GE=4b/b=4So GH//EF//BDTriangle HAD similar to triangle HLE So AG/GE=AD/El=2/3 and EH/ED=3/5=3/xSo x=5
Trace EF. BD meet AE and AF at P and Q.ΔABQ ~ QFD --> BQ = 2QD ΔAPD ~ EPB --> PD = 2BPBD = BQ + QD = 3 QD --> QD = 1/3x BD = BP + PD = 3 BP --> BP = 1/3xBy Thales’ theorem: EF // BD , EF = 1/2x.ΔBPG ~ ΔGEF --> PG/GE=BP/EF =(x/3)/(x/2)= 2/3 ΔQHD ~ ΔHEF --> QH/HF=QD/EF =(x/3)/(x/2)= 2/3 Then PG/GE = QH/HF = 2/3 ----- ( 1 )In the trapezoid EFQP, (1) --> GH // BD //EF or MH // BD //EF.ΔMEH ~ ΔBED --> MH/BD = GE/EP = 3/(3+2) = 3/5Then 3/x = 3/5 --> x=5.