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
Subscribe to:
Post Comments (Atom)
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