Saturday, July 1, 2017

Geometry Problem 1337: Parallelogram, Diagonals, Circle, Circumcircle, Tangent Line, Triangle, Congruence

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

Details: Click on the figure below.

Geometry Problem 1337: Parallelogram, Diagonals, Circle, Circumcircle, Tangent Line, Triangle, Congruence.


  1. I was originally going to prove tr. BGD congruent tr. ABD but a shortcut appeared during the process.

    Let Point H be the intersection of BD and CG
    and Let x = <CAD = <BCA and y = <ADB = <DBC (parallel lines)

    1. First <DEF = <CAD = x by the alternate segment theorem with tr ADE.
    2. From angle tracing <EDF = 180 - y and <DFE is y - x.
    3. Since <ACG inscribes the same arc as <DFE it also is y - x.
    4. Therefore from angle addition <BCG = <BCA + <ACG = x + (y - x) = y.
    5. Both <DBC and <BCG are y so tr BCH is isosceles and BH = CH
    6. But tr DGH is similar to BCH and therefore also isosceles and GH = DH.
    7. Adding the two pairs of equals segments together you get BD = CG.

  2. Let < FED = <GAC = @ and let < ACG = < EFD = €

    Then < CGF = @+€ = < EDG

    Hence the trapezoid BCDG is isoceles and hence concyclic and so BD = CG

    Sumith Peiris
    Sri Lanka

    1. Proving that AB = BG (which part has been added later on) is pretty straightforward since ABCD is a parellogram and BCGD is an isoceles trapezoid


    Let BD meet CG at K
    Since EF tangent to circle AED so ∠ (EAD)= ∠ (DEF)=u= ∠ (BCE)
    We have ∠ (EFG)= ∠ (ECG)= v ( both angles subtend to arc EG)
    And ∠ (ADE)= u+v ( external angle of triangle EDF)= ∠ (DBC)
    So ∠ (KBC)= ∠ (BCK)= u+v => triangle BKC is isosceles and KB=KC
    Similarly triangle KGD is also isosceles and KG=KD
    So BD=CG

  4. Draw orthogonal lines from B and c on AF to get the feet M and N respectively. Now consider the triangles BMD and CNG. In these we obviously have BM=CN and angles BMD=CNG. If we can show that also angles BDM=NGC, then the triangles are congruent and BD=CG.

    First we see NGC=FGC=FEC (as inscribed angles in the circumcircle of CEF), then we observe FEC=EDA(as angle between the chord and the tangent and inscribed angle in the circumcircle of AED)=BDM. Thus NGC=BDM, and we have proved all we needed.(by Thomas)

  5. FAE ~ FED lies FA:AE = FE:ED and FE = FA • ED / AE (1)
    ACG ~ AFE lies AC:CG = AF:FE and FE = CG • AF / AC (2)
    Now, from (1) and (2) follows
    FA • ED / AE = CG • AF / AC
    => ED / AE = CG / AC
    => ED • AC = AE • CG
    => 2ED • AC = 2AE • CG
    => BD • AC = AC • CG
    => BD = CG