Rhombi at the Incenter of a Triangle

AM

≈

4.11

IM

≈

4.11

IP

≈

4.11

AP

≈

4.11

IN

≈

2.77

NB

≈

2.77

BQ

≈

2.77

IQ

≈

2.77

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.