# 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

## Snapshots

## Details

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

## Permanent Citation

"Rhombi at the Incenter of a Triangle"

http://demonstrations.wolfram.com/RhombiAtTheIncenterOfATriangle/

Wolfram Demonstrations Project

Published: March 7 2011