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

Labels:
concurrent,
perpendicular bisector,
square,
triangle

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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