Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to enlarge the figure of problem 1013.

## Saturday, May 10, 2014

### Geometry Problem 1013: Square, Line through the Center, Perpendicular, Distance, Metric Relations

Labels:
center,
distance,
line,
metric relations,
perpendicular,
square

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Triangles EOA-FBO and GOD-HCO are congruent by ASA, making EO=FB and GO=HC. Therefore AE^2+(FB=EO)^2+GD^2+(CH=GO)^2=AO^2+OD^2=AD^2

ReplyDeleteBy symmetry,

ReplyDeleteAE=OF=OG=CH, BF=OE=OH=DG.

AE² + BF² + CH² + DG²

= AE² + OE² + OG² + DG²

= OA² + OD²

= AD²