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\angle AGH = \angle DHG in convex quadrilateral

Source: MEMO 2009, problem 3, single competition

October 1, 2009
geometryrhombusgeometric transformationhomothetygeometry proposed

Problem Statement

Let ABCD ABCD be a convex quadrilateral such that AB AB and CD CD are not parallel and AB\equal{}CD. The midpoints of the diagonals AC AC and BD BD are E E and F F, respectively. The line EF EF meets segments AB AB and CD CD at G G and H H, respectively. Show that \angle AGH \equal{} \angle DHG.
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