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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 879.
point F is 1/4 of the diagonal AC, drop a vertical from E and a horizontal from F gives us a base of 1 and a height of 3 for the triangle giving us an atan of 1/3 for half of the angle E. From here we find the tanE = 36.8698976..........
Draw EG//AB meeting AC in G, let /_FEG=α & /_GED=β, x = α + β . Further, tan(α)=1/3 and tan(β)=1/2 and so tan(x)=tan(α+β)=[tan(α)+tan(β)]/[1-tan(α)*tan(β)]=[1/3 + 1/2] = [1-1/6]=1. Hence, x=45°
EC=XDC=2Xbuild BDAC=sqrt(5)xAF=yFC=3y4y=sqrt(5)xy=sqrt(5)x/4=AFBD cut AC at QQF=sqrt(5)x/2-AF=sqrt(5)x/2-sqrt(5)x/4=sqrt(5)x/4FQ=AFDQ=sqrt(5)x/2DC/DQ=EC/FQ=2x/0.5sqrt(5)x=x/0.25sqrt(5)x==4/sqrt(5)FQD~ECDDFQ=DECF point on circle of ECD[can to prove it by "Reductio ad absurdum"]FCD=FED=45
http://img829.imageshack.us/img829/3494/22778697.pngAngle EKF=angle KED=90+45=135 degrees.If,AB=a ,is, FK=AK/2=a.sqrt(2)/4,EK=a/2 ,KD=a.sqrt(2)FK/EK=(sqrt(2))/2 (1) ,EK/KD=(sqrt(2))/2 (2)From (1),(2) , FK/EK= EK/KD so, the triangles EKF,EKD is similar ,therefore, angle FEK= angle EDK=ω, angle KED= angle KFE=y. But ω+y=45 ,so, angle x=45 degrees
http://img208.imageshack.us/img208/6976/problem897.pngDraw GFH //BASince AF/AC=1/4 so AH=BG=GE=1/4.ADTriangle DHF congruence to FGE (case SAS)So FD=FE and angle(GFE)+ angle (HFD)= 90 => angle (EFA)=90Triangle EFD is a right isosceles triangle => angle (FED)=45
From congruent triangles we see that FB = FE = FD so F is the centre of the circumcircle of Tr. BED. So < EFD = 2 < EBD = 90.Hence CDFE is cyclic and it follows that x = < FCD = 45.Sumith PeirisMoratuwaSri Lanka
Problem 879If the BD intersects AC at O then CO=2AF=2FO or AF=FO.But OF/OD=1/2=EC/CD so triangle ECD is similar with triangle FOD. Then <CDE=<ODF but <CDE+<EDB=45 or<ODB+<EDO=45 or <EDF=45=<ECF so ECDF is cyclic. Therefore <EFD=90 and <FED=45.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE