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.
Sunday, April 13, 2014
Geometry Problem 1004: Triangle with Squares, Perpendicular Bisectors, Concurrent Lines
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the perpendicular bisectors of B1C2, and A1B2 are concurrent at O.
ReplyDeleteM is midpoint of AB2. OA^2+OB2^2=2(OM^2+AM^2)=2(OM^2+CM^2)=OB1^2+OC^2
OA^2+OB^2=OC^2+OB1^2
OB^2+OC2^2=OA^2+OC1^2
OC^2+OA2^2=OB^2+OA1^2 therefore OA2=OC1