Geometry Problem. Post your solution in the comments box below.

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

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## Tuesday, July 14, 2015

### Geometry Problem 1135: Intersecting Circles, Secant, Parallel Line, 90 Degrees, Similar Triangles

Labels:
90,
intersecting circles,
parallel,
secant,
similarity,
triangle

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http://s13.postimg.org/7538xe84n/Pro1135.png

ReplyDeleteLet k= radius of circle O/radius of circle Q

Observe 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= k

And triangle ON’J similar to QNL( similar of right triangles) => JN’/LN=JG/LF= k

From above result will lead to JG/GB=JG/EF= LF/FD=BF/FH=GE/FH

So trinagles EGJ and HFE are similar ( case SAS)

Peter Tran,

ReplyDeleteI don't understand this ON'/ QN= OJ/OL= k

How did you come in this result.

Thank you

Observe that

DeleteM’C/MB=k=GC/BF=(M’C-GC)/(MB-BF)=M’G/MF=ON’/QN=k

Peter Tran

HI Peter,

ReplyDeleteI 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. QNL

DeleteSo JN’/NL= GN’/NF= k= (JN’+GN’)/(NL+NF)=GJ/LF= k

Similarly GB/FD= k => GJ/LF=GB/FD

Note that ∠ (GEF)= ∠ (EFD) and ∠ (EFL)= ∠ (FHE)+ ∠ (FEH)

But ∠ (GEF)- ∠ (EFL)= ∠ (LFD)=90

From above we can get ∠ (JEH)=90

Peter Tran

Name M intersecting of circle Q with HF.

ReplyDeleteConnect point J with C and B

point M with B and D

1. CG/GB=BF/FD (=CE/ED)=BC/BD

2. ∠OBQ= ∠QBD => ∠ OBC= ∠ BQD=> ∠ CJB= ∠BMD

3. 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