A property of incircle and circonscribed quadrilaterals.

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 
Proof :
draw AI and the perpendicular from B, intersecting in F.
AI intersects BC in D and BF intersects AC in E.
By construction, AI is symmetry axis of ABDE. Specifically AI is angle bissector of angle EDB
I is on the angle bissector of angle C, hence I is the excenter of triangle CDE.
Hence DE is tangent to incircle.

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.


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