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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
http://s13.postimg.org/7538xe84n/Pro1135.pngLet k= radius of circle O/radius of circle QObserve that following triangles are similar : CBD, CGE, EFD and OBQ ( case AA)And triangle COB similar to BQD ( case SSS)We have following ratios:BC/BD=BO/BQ=GC/GE=EF/FD=BG/FD=GC/BF=OM’/QM= ON’/QN=OJ/OL= kAnd triangle ON’J similar to QNL( similar of right triangles) => JN’/LN=JG/LF= kFrom above result will lead to JG/GB=JG/EF= LF/FD=BF/FH=GE/FHSo trinagles EGJ and HFE are similar ( case SAS)
Peter Tran,I don't understand this ON'/ QN= OJ/OL= k How did you come in this result. Thank you
Observe that M’C/MB=k=GC/BF=(M’C-GC)/(MB-BF)=M’G/MF=ON’/QN=kPeter Tran
HI Peter,I am not clear about this: JG/GB=LF/FD. Can you please explain this equivalence for me? Why angle JEH=90 degree?Thank You
Observe that ON’J similar to tri. QNLSo JN’/NL= GN’/NF= k= (JN’+GN’)/(NL+NF)=GJ/LF= kSimilarly GB/FD= k => GJ/LF=GB/FDNote that ∠ (GEF)= ∠ (EFD) and ∠ (EFL)= ∠ (FHE)+ ∠ (FEH)But ∠ (GEF)- ∠ (EFL)= ∠ (LFD)=90From above we can get ∠ (JEH)=90Peter Tran
Name M intersecting of circle Q with HF.Connect point J with C and B point M with B and D1. CG/GB=BF/FD (=CE/ED)=BC/BD2. ∠OBQ= ∠QBD => ∠ OBC= ∠ BQD=> ∠ CJB= ∠BMD3. Δ BCJ~ Δ BMD ( CG/GB=BF/FD =BC/BD; ∠BJC=∠BMD; JG⊥BC; FM⊥BD) MF/FD=JG/GB MF/FD=BF/FH JG/EF=GE/FH => Δ JGE~ ΔEFH ( case SAS)4. ∠GEF=∠EFD ∠EFM=∠FEH+∠JGE ∠JEH=∠QEF- (∠GEJ+∠HEF)= ∠EFD-∠EFM=90º