Perpendiculars from a Point on the Line between the Endpoints of the Angle Bisectors

P

PA''

≈

1.57

PB''

≈

2.93

PQ

≈

4.50

PA''

+

PB''

≈

4.50

Let ABC be a triangle. Let AA' and BB' bisect the angles CAB and ABC. Let P be a point on A'B' and PQ be perpendicular to AB. Let PB'' and PA'' be perpendicular to AC and BC. Then PQ = PA'' + PB''.