Wednesday, February 29, 2012

Problem 734: Triangle, Isosceles, Median, Parallel, Angle Bisector, Congruence, Midpoint

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

Click the figure below to see the problem 734 details.

Online Geometry Problem 734: Triangle, Isosceles, Median, Parallel, Angle Bisector, Congruence, Midpoint.

Posted by Antonio Gutierrez at 11:00 AM
Labels: , , , , , ,

7 comments:

  1. Peter TranFebruary 29, 2012 at 2:01 PM

    http://img812.imageshack.us/img812/4533/problem734.png
    Draw line NEL// BMC ( see sketch)
    Denote angles x, y, z, t as angles shown on the sketch
    Since M is the midpoint of BC => E is midpoint of NL
    Since DE//AB and E is the midpoint of NL =>
    DA=DL ( Thales’ theorem)
    Angles x=y=z=t (alternate angles and angles of isosceles triangle)
    Triangle DEB congruence to tri. DEL (case SAS)
    So angle (DBE)= angle (DLE)= angle (DCB)

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  2. PravinMarch 4, 2012 at 9:22 AM

    Draw MN ∥ ED (N on AC)
    MN/DE = AN/AD = NC/BD and
    ∠MNC = ∠BAD = ∠ABD = ∠BDE
    ∴ ∆DBE ~ ∆NCM (s.a.s)
    So ∠DBE=∠NCM=∠ACM

    ReplyDelete
  3. AjitMarch 6, 2012 at 7:37 PM

    Here's a slightly different proof:
    Extend BE to intersect DC in F. Consider tringle BFC with AEM as the transversal.
    By Menelaus, (BE/EF)*(FA/AC)*(CM/MA)=1 ignoring the –ve sign. But CM=MA and BE/EF =BD/DF by the angle bisector theorem since DE bisects /_ADF; hence
    BD/DF = AD/DF = AC/FA =(AD+DC)/(AD+DF) which gives AD^2=DF*DC or BD^2=DF*DC which implies that BD is tangent to the circumcircle of triangle BFC and thus /_DAE=/_C by the alternate segment theorem.
    Ajit

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  4. PravinMarch 8, 2012 at 8:28 PM

    Yet another solution to Problem 734:
    For convenience denote AD=BD=d, ∠DBE=x
    Note ∠EDC=A=∠BDE
    Draw MN∥DE(N on AC)
    MN∥AB and MN=(1/2)AB
    DE/MN = AD/AN = 2d/b
    Follows DE = AD.MN/AN=AD.2MN/2AN=AD.AB/AC = dc/b
    Also in ∆BDE, by Cosine Rule
    BE²=BD²+DE²-2BD.DE cos A
    =d²+(d²c²/b²)-2(d²c/b)cos A
    =[d²/b²][b²+c²-2bc.cos A]=[d²/b²].a²
    So BE = da/b
    BE:BD:DE=(da/b):d:(dc/b)=a:b:c
    Thus ∆DEB ∼ ∆ABC
    Hence ∠DBE = ∠ACB

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  5. WFungApril 5, 2012 at 12:05 AM

    We know that ∠BAD = ∠BDE.
    Hence we just need to prove that 2 prop. sides for similar triangles to complete the entire proof.
    By menelaus, |(BM/MC)*(CA/AD)*(DH/HB)|=1
    But BM = MC , (CA/AD)*(DH/HB)=1
    Also ∆ABH ~ ∆HED, (HB/DH)=(AB/ED)
    Combining and we get (CA/AB)=(BD/ED).
    Q.E.D.

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  6. zs359142279May 13, 2012 at 1:07 PM

    I used menelaus too....
    ∵BE//AB
    ∴∠ABD=∠BDE (alternate interior angles theorem)
    ∵AD=BD
    ∴∠ABD=∠BAD (base angles theorem)
    ∴∠BDE=∠BAD
    Goal: Prove that △ABC∼△DEB so that ∠DBE=∠ACB
    AM and BD intersect at K
    ∵△AKB∼△EKD
    ∴DE/AB=DK/BK
    (Goal: Prove that DK/BK=BD/AC=AD/AC)
    BM/CM*AC/AD*DK/BK=1 (Menelaus' theorem)
    ∵BM=CM (Def. of medians)
    ∴AC/AD*DK/BK=1
    AC/AD=BK/DK
    AD/AC=DK/BK=BD/AC
    ∴DE/AB=BD/AC
    ∵∠BDE=∠BAD
    ∴△ABC∼△DEB
    ∴∠DBE=∠ACB

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  7. Sumith PeirisFebruary 15, 2017 at 9:11 AM

    If X is midpoint of AC, then MX//DE//AB and
    MX = c/2

    If AD = p = BD and DE = q

    (b/2)/p = (c/2)/q so p/q = b/c
    So triangles ABC and DBE are similar and so the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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