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The figure below shows the squares ABCD and DEFG (E on CD). GE extended meets BC at H. CM is parallel to HG (M on FG). Prove that (1) AH =HM; (2) angle AHM = 90 degrees.

See also: Art of problem 1312

Home | Geometry | Problems | All Problems | Open Problems | Visual Index | 1311-1320 | Triangle | Square | Parallel lines | Perpendicular line | Angle | Congruence | View or Post a solution | by Antonio Gutierrez