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 |

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.