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

Labels:
diagonal,
intersection,
parallel,
parallelogram

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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.