Tuesday, July 14, 2015

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

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1135: Intersecting Circles, Secant, Parallel Line, 90 Degrees, Similar Triangles.

Posted by Antonio Gutierrez at 3:59 PM
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7 comments:

  1. Peter TranJuly 17, 2015 at 9:44 PM

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

    Let 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)

    ReplyDelete
  2. AnonymousJuly 24, 2015 at 8:49 AM

    Peter Tran,

    I don't understand this ON'/ QN= OJ/OL= k
    How did you come in this result.

    Thank you

    ReplyDelete
    Replies
    1. Peter TranJuly 25, 2015 at 10:25 PM

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

      Peter Tran

      Delete
  3. AnonymousJuly 25, 2015 at 4:12 PM

    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

    ReplyDelete
    Replies
    1. Peter TranJuly 25, 2015 at 11:28 PM

      Observe that ON’J similar to tri. QNL
      So 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

      Delete
  4. AnonymousJuly 27, 2015 at 8:24 AM

    Name M intersecting of circle Q with HF.
    Connect 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

    ReplyDelete
  5. Sumith PeirisOctober 26, 2018 at 1:30 AM

    Let JG = u, FH = v, CG = w, GE = x = BF, BG = y, EF

    u2 = wy ….(1)

    v2 = xz …. (2)

    w/x = y/z ….(3) since Tr.s CGE and EFD are similar

    So (u2 / y) / x = y /( v2 / x) from which,
    u/x = y/v and since < JGE = < EFH,
    Tr.s JGE & EFH are similar

    Now
    < JEH = < GEF - < GEJ - < HEF
    = < GEF - < GEJ - < GJE
    = 180 - < EGB - < GEJ - < GJE
    = 90 considering Tr. JGE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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