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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 962.
Let BE=mAE, BD=nCD. Let S(ABC) denote the area of ABC. From Menelaus theorem, nCF=(m+1)EFmAF=(n+1)DFS₁ = S(ADC) = 1/(n+1) S(ABC)S₃ = S(AEC) = 1/(m+1) S(ABC)S₂ = S(BEF) = n/(m+n+1) S(BEC) = mn/[(m+1)(m+n+1)] S(ABC)S₄ = S(BDF) = m/(m+n+1) S(BDA) = mn/[(n+1)(m+n+1)] S(ABC)Thus, S₁×S₂ = S₃×S₄ = mn/[(m+1)(n+1)(m+n+1)] S(ABC)²
Extend BF and cut AC at GBy Ceva, (BD/DC)*(CG/GA)*(AE/EB) = 1 So BD*CG*AE = BE*AG*DCS1*S2 =ΔADC*ΔBEF=(DC/BC)*ΔABC * (BE/BA)*ΔBAF=(DC/BC)*ΔABC * (BE/BA)*(BF/BG)*ΔBAG=(DC/BC)*ΔABC * (BE/BA)*(BF/BG)*(AG/AC)*ΔABC=(DC/BC)*(BE/BA)*(AG/AC)* [ΔABC]^2*(BF/BG)=(DC*BE*AG)/(BC/BA/AC) * [ΔABC]^2*(BF/BG)=(AE*BD*GC)/(AB/BC/AC) * [ΔABC]^2*(BF/BG)=(AE/AB)*(BD/BC)*(GC/AC) * [ΔABC]^2*(BF/BG)=(AE/AB)*ΔABC * (BD/BC)*(BF/BG)*(GC/AC)*ΔABC=(AE/AB)*ΔABC * (BD/BC)*(BF/BG)*ΔBGC=(AE/AB)*ΔABC * (BD/BC)*ΔBFC=S3*S4
http://imagizer.imageshack.us/v2/800x600q90/42/epgz.pngLet ED cut AC at KLet L, M are projection of E and D over ACNote that (K,E,H,D)=-1And HE/HD=KE/KD=EL/DM….. (1)But HE/HD= h2/h4We have S2/S4=h2/h4 …. ( triangles have the same base)And S3/S1=EL/DM…. (triangles have the same base)Per (1) we have h2/h4=EL/DM => S2/S4=S3/S1 => S1.S3=S3.S4
Extend ED to AC at X. ED cuts BF at Y. S1/S3=perpendicular from D to AC/perpendicular from E to AC=XD/XE, and S4/S2=perpendicular from D to BF/perpendicular from E to BF=DY/EY.Applying Menelaus to BED and transversal XAC, we get that AE/AB*BC/DC*DX/EX=1Applying Ceva to BED and point F, we obtainAE/AB*BC/DC*YD/YE=1From these two equations we see that S1/S3=DX/EX=YD/YE=S4/S2