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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 1004.
the perpendicular bisectors of B1C2, and A1B2 are concurrent at O.M is midpoint of AB2. OA^2+OB2^2=2(OM^2+AM^2)=2(OM^2+CM^2)=OB1^2+OC^2OA^2+OB^2=OC^2+OB1^2OB^2+OC2^2=OA^2+OC1^2OC^2+OA2^2=OB^2+OA1^2 therefore OA2=OC1