Geometry Problem
Level: Mathematics Education, High School, College.
Click the figure below to see the complete problem 614.
Monday, June 6, 2011
Problem 614: Parallelogram, Diagonals, Intersecting Lines, Parallel Lines
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Consider ∆BED and transversal CMF
ReplyDeleteBy 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
Another way to solve using similar triangles.
ReplyDeletehttp://img687.imageshack.us/img687/6279/problem614.png
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)
HE/HA=HB/HK=BE/AK=a.b/(a+b)^2
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
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 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.
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
ReplyDelete1)EC/BE*(BF/FD=BE/AD)*DP2/CP2=1
2)BE/EC*(CG/AG=EC/AD)*AP1/BP1=1
It follows that DP2/CP2=AD/EC and AP1/BP1=AD/BE.
From Aubel II we know that
AH/HE=(AG/CG=AD/EC)+(AP1/BP1=AD/BE)
DM/ME=(FD/BF=AD/BE)+(DP2/CP2=AD/EC)
We have shown that AH/HE=DM/ME so we are done.