Monday, June 6, 2011

Problem 614: Parallelogram, Diagonals, Intersecting Lines, Parallel Lines

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

Click the figure below to see the complete problem 614.

Problem 614: Parallelogram, Diagonals, Intersecting Lines, Parallel Lines


  1. Consider ∆BED and transversal CMF
    By Menelau:
    (EM/MD) (BC/CE) (DF/FB) = 1
    Consider ∆CEA and transversal BHG
    By Menelau:
    (EH/HA) (AG/GC) (CB/BE) = 1
    Note that
    (BC/CE) = (AD/EC) = (AG/GC) and
    (DF/FB) = (AD/BE) = (CB/BE)
    It follows
    (EM/MD) = (EH/HA)
    Hence HM ∥AD

  2. Another way to solve using similar triangles.

    Let CF cut AD at H, BG cut AD at K
    Let a=BE, b=EC
    Using similar triangles we have
    BE/DK=GE/GD=b/(a+b) so DK=a/b*(a+b)

    With the same way as above we have AH=b/a*(a+b)
    And EM/MD=EC/DH=a.b/(a+b)^2

    So EM/MD=HE/HA and HM //AD

    Peter Tran

  3. There is no need for E to be on BC.

    The dual of Pappus hexagon theorem says that if lines p, p’, p’’ concur, and lines q, q’, q’’ concur, and if you call (I'll call a^b the intersection point of lines a and b)
    r the line joining p^q’ with p’^q,
    r’ the line joining p’^q’’ with p’’^q’,
    r’’ the line joining p^q’’ with p’’^q,
    then lines r, r’, r’’ concur or are parallel.

    In this problem lines FBD (p), FCM(p’), FAH (p’’) concur in F; lines GCA (q), GBH (q’), GDM (q’’) concur in G then line BC (r), HM (r’) and line AD (r’’) concur or are parallel. Since AD//BC then also HM//AD//BC.

  4. CH cuts AB at P1 and BM cuts CD at P2. Applying Ceva twice, once to M in triangle BCD, and once to H in BAC, we obtain
    It follows that DP2/CP2=AD/EC and AP1/BP1=AD/BE.
    From Aubel II we know that
    We have shown that AH/HE=DM/ME so we are done.