Friday, June 1, 2012

Problem 759: Equilateral Triangle, Transversal, Trisection of sides, Congruence, Angle

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

Click the figure below to see the problem 759 details.

Online Geometry Problem 759: Equilateral Triangle, Transversal, Trisection of sides, Congruence, Angle.

Posted by Antonio Gutierrez at 8:51 PM
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9 comments:

  1. WFungJune 1, 2012 at 11:09 PM

    By Menelaus theorem on triangle ABC, (BD/DC)*(CF/FA)*(AE/EB)=1.
    Hence CF/FA = 4/1. Therefore FA:AC=1:3.
    Since ABC is an equilateral triangle, FA = (1/3)AC = (1/3)AB = AE
    So triangle AEF is an isosceles triangle. Angle EAF = 120, hence x = 30.

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  2. Peter TranJune 1, 2012 at 11:10 PM

    please disregard my previous solution. new Solution below include minor typo corrections.
    Apply Menelaus’s theorem to triangle ABC with secant FED
    DB/DC x FC/FA x EA/EB=1 => ½ x FC/FA x ½= 1
    So FA/FC= ¼ => CA/CF= ¾
    From A draw AH //FD ( H on CB)
    We have CA/CF=CH/CD= ¾ => CH=3/4.CD=3/4 .2/3 .CB= ½. CB
    AH will be an angle bisector and altitude of triangle ABC, so x= angle(HAC)= 30

    ReplyDelete
  3. PravinJune 2, 2012 at 4:09 AM

    Draw EG ∥ DC (G on AC)
    ∆s GDC and AEG are equilateral and
    BEGD is a parallelogram
    1/2 = EG/DC = FG/FC
    G is the midpoint of FC
    FG = GC = EB = DG
    ∆FDC is rt angled at D
    x = 30°

    ReplyDelete
  4. PravinJune 2, 2012 at 4:39 AM

    AE = BD = a/3 where AB = BC = AC = a
    So BE = CD = 2a/3
    Apply Menelau's Theorem to ∆ABC
    and transversal DEF
    (BE/EA).(AF/FC).(CD/DB) = 1 numerically
    Easy to see BE/EA = 2 = CD/DB
    Follows AF/FC = 1/4, AF = a/3 = AE
    So x = ∠AFE = ∠AEF = (1/2)∠EAC = 30°

    ReplyDelete
  5. CiPNovember 27, 2012 at 8:57 AM

    Let G be the midpoint of [BE].(Draw a picture yourself)
    From BG/BA=BD/BC =1/3 follow that triangles BDG and BAC are similar; moreover DG/CA=1/3.
    Now, in triangle BDE, median DG=BE/2- half of side of base, or that suffice that angle BDE is right. In fact, because triangle BDG is equilateral, follow that triangle EBD (in that order of vertices) is of type 30-60-90.
    The rest is simple.

    Visit also http://ogeometrie-cip.blogspot.com/

    ReplyDelete
  6. kalpitFebruary 23, 2013 at 7:08 AM

    Let BD = AE = a . Draw DG // AB intersecting AC at G. now DC = BE = 2a. In tr.ABC, by BPT, GC = 2a and AG = a.
    now tr. DGC is equilateral. so DG = 2a . In tr. FGD and tr.FAE are similar. so FA = a. Now FA = AE = a. So ∠FEA = ∠AFE = x. ∠ FEA = ∠ GDF = x. In tr. DFG ∠DFG + ∠ GDF = ∠DGC. 2x = 60° or x = 30°

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  7. Μιχάλης ΝάννοςOctober 10, 2013 at 6:27 AM

    http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27308&p=133642

    ReplyDelete
  8. Sumith PeirisSeptember 20, 2015 at 4:53 AM

    Easily BED is a 30-60-90 Tr. Having BE = 2.BD and the included angle 60. So < BED = 30 and hence x = 30.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  9. MedFebruary 25, 2019 at 4:55 AM

    posons AB = R et AH la hauteur issue de A.
    Alors : BD = R/3 ; BH = R/2 et BE = 2R/3
    BD/BH=2/3 et BE/BA=2/3, donc BD/BH= BE/BA
    D’où : (ED) parallèle à (AH) et x= 30°.
    English translation :
    let AB = R and AH be the height of A.
    Then: BD = R / 3; BH = R / 2 and BE = 2R / 3
    BD / BH = 2/3 and BE / BA = 2/3, so BD / BH = BE / BA
    Hence: (ED) parallel to (AH) and x = 30 °.

    ReplyDelete

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