## Wednesday, December 19, 2012

### Problem 834: Parallelogram, Diagonal, Similarity, Geometric Mean, Mean Proportional

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

Click the figure below to see the complete problem 834.

1. Note that triangle BFA similar to triangle GFC ( case AA)
so BF/FG=FA/FC
Triangle BFC similar to triangle EFA ( case AA)
so FE/BF=FA/FC => BF/FG=FE/BF => BF=GM(FG, FE)

2. From ΔAEF~ΔCBF,
AF/FC = FE/BF

From ΔABF~ΔCGF,
AF/FC = BF/FG

Hence,
BF/FG = FE/BF
BF^2 = FG×FE
BF = GM(FG,FE) = √(FG×FE)

3. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 20, 2012 at 3:56 AM

Daca notam cu P intersectia ,paralelei prin D la BG ,vom obtine: paralelogramul BFDP, AF = PC=>AP=FC.In continuare aplicam teorema fundamentala a asemanarii in triunghiurile CFG,PD||FG si respectiv APD,EF||PD =>PD/FG=CP/CF=AF/CF=AF/AP=FE/PD=>
PD=GM(FG,FE),qed.

4. triangle CGB similar to triangle ABE so CG/AB=CB/AE. But CG/AB=FG/BF, and CB/AE=BF/FE, so substituting these values in, we get FG/BF=BF/FE, or BF^2=FG*FE

5. EX is parallel to CD such that X is on AC. Then BF/FG=AB/CG=AE/BC=AE/AD=EX/CD=EX/AB=FE/BF, second equality due to fact that triangles ABE and CGB are similar. Result follows from BF/FG=FE/BF.