Geometry Problem

Level: Mathematics Education, High School, College.

Click the figure below to see the complete problem 616.

## Thursday, June 9, 2011

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

Labels:
diagonal,
intersecting lines,
parallel,
parallelogram

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http://img827.imageshack.us/img827/9192/problem616.png

ReplyDeleteLet 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

HB/HL=BM/CL=a/b

So HB/HL= BE/EC

And HE//CD ( Thales theorem on parallel lines)

Peter Tran

There is no need for E to be on BC.

ReplyDeleteThe 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.

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

ReplyDeleteBE/EC*(CO/AO=1)*AP1/BP1=1

EC/BE*(BO/DO=1)*DP2/CP2=1

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.