Thursday, September 8, 2016

Geometry Problem 1256: Parallelogram, Angle Bisector, Measurement

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 1256.

Geometry Problem 1256: Parallelogram, Angle Bisector, Measurement.

Posted by Antonio Gutierrez at 8:17 PM
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8 comments:

  1. Peter TranSeptember 8, 2016 at 9:14 PM

    https://goo.gl/photos/YYzh4r9rVdU65h2p7

    Let CE meet BA extended at N
    We have ∠ (BCE)= ∠ (ECD)= ∠ (CED)= ∠ (ANE)= u ( see sketch)
    So triangle CDE is isosceles and CD=ED=FG=AB= 28
    Triangle GFN is isosceles so FG=FN= 28
    Triangle CBN is isosceles so x= BC= BN= 15+28= 43

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  2. APOSTOLIS MANOLOUDISSeptember 8, 2016 at 9:41 PM

    Problem 1256
    If CE intersects AB at K then <BCE=<ECD=<CED=<AEK=<AKE .So AB=CD=DE=FG=15+13=28.
    And AE=AK=y then triangles KAE and KFG are similar.So AE/FG=KA/KF or y/28=y/y+13 or
    y+13=28 so y=15.But x=BC=AD=y+28=15+28=43.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  3. AnonymousSeptember 8, 2016 at 10:33 PM

    As [FG]=[ED], [AE]=[GX], being X the intersection of FG and DC.

    Since [FBCX] is a parallelogram as well, ^GXC= 180-2(^GCD); So, ^CGX=^GCD and [GX]=[XC]=[BF]=15.

    Let P be the intersection of BA and EC. Let [AP]=y.

    Since [PAE], [PFG] and [PBC] are all similar, we have:

    y/(y+13)=15/(x-15); x= 30 + (13.15/y)

    And (y+28)/y= (30+ (13.15/y))/15; y=15

    So, x= 30+13 = 43.


    DL06

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  4. APOSTOLIS MANOLOUDISSeptember 8, 2016 at 10:58 PM

    Problem 1256
    Solution 2
    If CE intersect the AB in K then <BCE=<ECD=<DEC=<AEK=<AKE=<FGK so AB=CD=ED=FG=FK=15+13=28 and BC=BK. So AK+AF=FG or AK+13=28 or
    AK=15.Therefore x=BC=BK=15+13+15=43.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  5. Sumith PeirisSeptember 9, 2016 at 2:54 PM

    Let FG meet CD at H.
    Tr. CGH is isoceles.
    So x - 28 = 15 and hence x = 43

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. UnknownSeptember 11, 2016 at 5:59 AM

    We use Barycentric Coordinates.
    Let A(1,0,0), B(0,1,0), C(0,0,1)
    B + D = A + C (Because ABCD is parallelogram).
    Hence, D = A – B + C = (1,-1,1)
    F = (15,13,0)/28 (F is on AB, in ratio 13:15)
    E is on AB. Let E = A+D(1-q) for some scalar q. Using definition of D, E = (1,-q,q).
    Line CE has equation qx + y = 0
    [Every line in Barycentric coordinates is in the form ux + vy + wz = 0 for some constant u, v, w. The coefficients are determined by subbing in (x,y,z) and solving for u, v, w.]
    G + E = F + D as DEFG is a parallelogram. Since we know E, F and D, we can rearrange this equation to make G = (15/28, q – 15/28, 1 – q)
    G is on line CE, so: 15q/28 + q – 15/28 = 0. This means that q = 15/43. G = (15/28, -225/1204, 28/43)
    Line GD has equation 559x + 139y – 420z = 0
    Make point H which is intersection of GD and BC. This implies that x = 0. H = (0, 420/559, 139/559)
    It is obvious that 13H/28 + 15D/28 = G
    This means that HG:DG = 15:13
    By the angle bisector theorem, it is obvious that HG:DG = HC:CD.
    Hence, HC = 420/13
    It is also obvious that 420B/559 + 139C/559 = H
    Hence, BH:CH = 139:420
    This means that HC:CB = 420:559
    Finally, CB = (420/13)*(559/420) = 43

    Not the cleanest way of doing it, but for fist-brains like me who can't find simple answers, this is epic.

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    Replies
    1. UnknownSeptember 11, 2016 at 6:13 AM

      TYPO ALERT: When I said "E = A+D(1-q)", I actually meant to say "E = A(1-q)+Dq"

      Delete
  7. MarcoJune 20, 2024 at 1:56 AM

    <CED=<BCE (alt <s BC//DE)
    So, <CED=<ECD
    FG=ED=DC=BA=28
    Join BE and let the intersection point of BE and FG be H
    FH/AE=15/28
    FH=15AE/28
    HG/BC=13/28
    HG=13x/28
    15AE/28+13x/28=FH+HG=FG=28
    15(x-28)+13x=784
    28x-420=784
    x=43

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