The Collinearity of Feet of Perpendiculars to Bisectors - pending

Also known as:
The Midline Property of Angle Bisectors

Here is the statement:
When you drop perpendiculars from a vertex to the internal and external bisectors of the other two angles, the four feet always lie on a single straight line (the midline of the triangle).

Proof:
Internal angle bisector of B is the line which sits midway between AB and BC.
AB reflected in this will overlap with BC.
If you take A's reflection in this which maps to A', A' will lie somewhere on BC.
And the internal bisector of B will be the perpendicular bisector of AA'.
Let Q be the midpoint of AA'.
Similarly A'' is A's reflection in angle B's external bisector. A'' will lie on BC.
We can again show that if AA'' has P as its midpoint then BP is the perpendicular bisector AA''.

Let B be (0,0). A' = (x1,0), A'' = (x2,0), A = (x1,y1)
y co-ordinates of P and Q will be same  = y1/2
Midpoints of AB and AC will also have the same y co-ordinate.

So P,Q and midpoints of AB,AC are colinear and parallel to BC.

Same goes for A's reflection in C's internal/external bisectors.

So feet of perpendiculars from A to B,C 's angle bisectors are colinear and parallel to BC.
H.P.