## Monday, January 20, 2014

### Geometry Problem 962. Triangle, Two Cevians, Areas, Equal Products

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

Click the figure below to see the complete problem 962.

1. Let BE=mAE, BD=nCD.
Let S(ABC) denote the area of ABC.

From Menelaus theorem,
nCF=(m+1)EF
mAF=(n+1)DF

S₁ = 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

2. Extend BF and cut AC at G
By Ceva, (BD/DC)*(CG/GA)*(AE/EB) = 1
So BD*CG*AE = BE*AG*DC

S1*S2
=(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

3. http://imagizer.imageshack.us/v2/800x600q90/42/epgz.png

Let ED cut AC at K
Let L, M are projection of E and D over AC
Note that (K,E,H,D)=-1
And HE/HD=KE/KD=EL/DM….. (1)
But HE/HD= h2/h4
We 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

4. 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=1

Applying Ceva to BED and point F, we obtain
AE/AB*BC/DC*YD/YE=1

From these two equations we see that S1/S3=DX/EX=YD/YE=S4/S2