Given a triangle ABC, its incircle centered in I, touching AC in U and BC in V
|AI, UV and the perpendicular to AI from B are concurrent|
Consider hexagon AUEDVB circumscribed to circle.
(degenerated as U and V are both vertices and touching points)
The Brianchon theorem says that diagonals of a circumscribed hexagon are concurrent.
Hence AD, BE, UV concurrent. QED.
The reciprocal property is interresting :
|A circumscribed quadrilateral with perpendicular diagonals is symmetric from one of these diagonals|
Let a circumscribed quadrilateral ABCD, we may fix AB, angle A and angle B, at least one of them being acute.
This fixes the incircle and triangle PAB.
We'll search the fourth side CD so as to have AC_|_BD.
Point F intersect point of AC and BD is then on the circle with diameter AB.
The quadrilateral being circumscribed, the Brianchon theorem gives F on UV.
F is then intersect point of UV with circle with diameter AB, that is just two points F and F'.
From the direct property above, AF is angle bissector of angle A. (and BF' of angle B). QED.