Proposed Problem
Click the figure below to see the complete Routh's Theorem 2 about Triangle, Cevians, Ratio, Areas.
See also:
Routh's Theorem 2
Level: High School, SAT Prep, College geometry
Sunday, January 10, 2010
Routh's Theorem 2: Triangle, Cevians, Ratio, Areas
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http://i001.radikal.ru/1001/e6/9bebb0652714.gif
ReplyDeletehttp://s005.radikal.ru/i209/1001/9d/465e02317c25.gif
EF || BD
There are some errors in my previous comment. Here is the corrected comment.
ReplyDeleteLet line BF cut AC at H
Let S2= area of Tri. AFC ; S3= area of tri. BEC and S4=area of tri. BDA and S=area of tri.(ABC)
1. Note that Area(AFC)/Area(ABC) =FH/BH ( both triangles have same base)
2. Apply Van Aubel theorem for Cevians CC’, AA’ and BH in triangle ABC
We have BF/FH =BC’/AC’ +BA’/A’C =n+1/k
3. Area(AFC)/Area(ABC)=S2/S =FH/(BF+FH)=k/(nk+k+1)
4. Similarly S3/S= n/(nm+n+1) and S4/S=m/(km+m+1)
5. Area S1= S-S2-S3-S4
6. S1=S*[1-(k)/(nk+k+1)-(n)/(mn+n+1)-(m)/(km+m+1) ]
7. Develop this we will get the result
Peter Tran
http://img839.imageshack.us/img839/6070/rouththeorem2.png
ReplyDeleteAttached is the sketch for above problem
Peter Tran