## Sunday, January 10, 2010

### Routh's Theorem 2: Triangle, Cevians, Ratio, Areas

Proposed Problem
Click the figure below to see the complete Routh's Theorem 2 about Triangle, Cevians, Ratio, Areas.

Routh's Theorem 2
Level: High School, SAT Prep, College geometry

EF || BD

2. There are some errors in my previous comment. Here is the corrected comment.
Let 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

3. http://img839.imageshack.us/img839/6070/rouththeorem2.png
Attached is the sketch for above problem

Peter Tran