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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 928.
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.
By affine transformation to any isosceles triangle, all the properties in the problem is preserved.After the transformation, S1=S4 ; S2=S3 ; S5=S6As affine preserved area ratio, the proof is completed.
S6=S5 => h1=h2 (h1=AA', h2=CC' alt) => h3=h4 (h3=FF', h4=DD') because Right tr AA'B~ tr FF'B and BCC'~BFF'=> S2=S3 => S1=S4