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: Triangle, Cevians, Ratio, Areas.
See also:
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

    Attached is the sketch for above problem

    Peter Tran