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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1033.
Let S(ABC) be the area of ABC. S(CDE) = S(ABF) = 2S(ACDE) = 11+2 = 13S(OCDE) = 13/2S(OCE) = 13/2 - 2 = 9/2S(BCE) = 9S(AEF) = S(BCD) = 4S(ACEF) = 11+4 = 15S(AOEF) = 15/2S(AOE) = 15/2 - 4 = 7/2S(ABE) = 7S(ABC) = 9+7-11 = 5
AC=FD, CE=BF, and AE=BD, so [BFD]=[ECA]=11. [ABC]=2+4+11-[ACDF]=2+4+11-2[ACF]=2+4+11-2(2+4)=5