Rhombi at the Incenter of a Triangle

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Let ABC be a triangle and let I be the intersection of the angle bisectors. Let MN be parallel to AB and through I, with M on CA and N on BC. Let P and Q be points on AB such that IP and IQ are parallel to CA and BC, respectively. Then AMIP and IQNB are rhombi.

Contributed by: Jay Warendorff (March 2011)
Open content licensed under CC BY-NC-SA


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See problem 5.44 in V. Prasolov, Problems in Plane and Solid Geometry, Vol. 1,Plane Geometry [PDF], (D. Leites, ed. and trans.).



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