Thursday, June 9, 2011

Problem 616: Parallelogram, Diagonals, Intersecting Lines, Parallel Lines

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

Click the figure below to see the complete problem 616.

Problem 616: Parallelogram, Diagonals, Intersecting Lines, Parallel Lines



    Let BG cut CD at L and CF cut AB at M
    Let a=BE , b=EC and c=AB=CD

    Triangle(EGC) similar to Triangle (DGA);
    Triangle(BGA) similar to Triangle (LGC) case AA
    So GC/GA=b/(a+b)=CL/AB and CL=b.c/(a+b)

    With the same way as above BM=a.c/(a+b)

    Triangle (BHM) similar to Triangle (LHC) case AA
    So HB/HL= BE/EC
    And HE//CD ( Thales theorem on parallel lines)

    Peter Tran

  2. 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 FAE (p), FDB (p’), FHC (p’’) concur in F; lines GHB (q), GAC (q’), GDE (q’’) concur in G then line AB (r), line DC (r’) and line HE (r’’) concur or are parallel. Since AB//DC then also HE//AB//DC.

  3. CF cuts AB at P1 and BG cuts CD at P2. Let O be the center of the parallelogram. By Ceva in triangle ABC with point F and in triangle BCD with point G, we obtain
    Therefore AP1/BP1=EC/BE and DP2/CP2=BE/EC.
    Because AB=CD, BH/HP2=BP1/CP2=BE/CE so EH is parallel to CD.