Monday, November 28, 2016

Geometry Problem 1290 Triangle, Internal Angle Bisector, Median, Parallel, Measurement, Metric Relations

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

Click the figure below to view more details of problem 1290.

Geometry Problem 1290 Elements: Triangle, Internal Angle Bisector, Median, Parallel, Measurement, Metric Relations.

Posted by Antonio Gutierrez at 12:45 PM
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9 comments:

  1. Peter TranNovember 28, 2016 at 6:13 PM

    https://goo.gl/photos/SxYFtYyAsxydH45r7
    Let ME extended meet BC at F
    Draw AN //ME . N is on BC extended ( see Sketch)
    Observe that ∠BEF=∠BFE= u => triangle EBF is isosceles
    ∠BAN=∠ANB=u => triangle ABN is isoceles
    Since M is the midpoint of AC => F is the midpoint of NC=> FN=FC
    AB=BN=FN+FB=FC+x
    BC=FC-FB =FC-x
    So AB-BC= FC+x-(FC-x)=2x= 5 => x= 2.5

    ReplyDelete
  2. Sumith PeirisNovember 28, 2016 at 8:47 PM

    With the usual notation for triangle ABC,

    DC = ab/(a+c) and so
    MD = b/2 - ab/(a+c) = (b/2)(c-a)/(c+a)

    Now ME//BD so
    (c-x)/x = (b/2)/MD from which
    c/x = 1 + (b/2)/MD
    = 1 + (c+a)/(c-a) = 2c/(c-a)

    Hence x = (c-a)/2 = 5/2 = 2.5

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Pradyumna AgasheNovember 28, 2016 at 11:59 PM

    Extend AB to point F, such that BF=BC
    Angle AFC= Angle ABC/2, hence FC is parallel to ME.
    Since M is midpoint of AC,E must be midpoint of AF, AE=(AB+BC)/2. BE=AB-AE=(AB-BC)/2
    BE=5/2

    ReplyDelete
  4. APOSTOLIS MANOLOUDISNovember 29, 2016 at 1:47 AM

    Problem 1290
    Draw AP //BD//ME . P is on AB extended.Is <BCP=<DBC=<DBA=<BPC, then
    BP=BC and AM=BC so AE=EP or AB-BE=BD+BE or AB-BC=2.BE or x=2.5.
    MANOLOUDIS APOSTOLIS4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE

    ReplyDelete
  5. Stan FulgerNovember 30, 2016 at 12:27 AM

    Extend ME to cut BC at F, construct the parallelogram CFEG; due to AM=CM, ME passes through the midpoint of AG and, as it is the angle bisector of <AEG, tr. AEG is isosceles and AE=GE(=CF). With BF=BE, we are easily done.

    ReplyDelete
  6. AnonymousNovember 30, 2016 at 6:53 AM

    Let AB = c, BC = a and AC = b
    we have DC = ab/a+c , AD = bc/a+c and AM = b/2
    Since triangles AME and ADB are similar
    AE/AM = AB/AD
    => c-x/(b/2) = c/(bc/a+c)
    => 2c-2x = a+c
    => x = c-a/2 = 2.5

    ReplyDelete
  7. c.t.e.oDecember 1, 2016 at 8:26 AM

    Draw CF Perpendic to BD, MN //CF
    => MN = 1/2 AE = 2.5 (MG middle line of AFC)
    => MG=EB (MGBE parallelogram)

    ReplyDelete
    Replies
    1. Antonio GutierrezDecember 1, 2016 at 12:47 PM

      To c.t.e.o
      Where are points F and G?
      MN is not necessary, use MG. Thanks

      Delete
  8. c.t.e.oDecember 1, 2016 at 1:29 PM

    G point CF meet BD, F on AB (Tr BFC isoceles)
    MN is also middle line of AFC, a way to prove equality of tr MNE and DCB
    Thanks

    ReplyDelete

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