Thursday, September 26, 2013

Geometry Problem 928: Triangle, Midpoint, Median, Cevian, Concurrency, Area

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 928.

Online Geometry Problem 928: Triangle, Midpoint, Median, Cevian, Concurrency, Area

3 comments:

  1. Denote the area of ΔABC by S(ABC).

    Obviously, S(AME)=S(CME). i.e. S5=S6.

    By Ceva's theorem, BF/AF=BD/CD. i.e. FD//AC.
    Thus, S(FAC)=S(DAC), which implies S1=S4.

    Lastly, S(BAM)=S(BCM), which implies S1=S4.

    Hence, S1+S3+S5 = S2+S4+S6 = S/2.

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  2. By affine transformation to any isosceles triangle, all the properties in the problem is preserved.
    After the transformation,
    S1=S4 ; S2=S3 ; S5=S6
    As affine preserved area ratio, the proof is completed.

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  3. http://www.mathematica.gr/forum/viewtopic.php?f=22&t=40193&p=187258#p187258

    ReplyDelete