Incircle

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.

 

Home Arithmetic Geometric Misc Topics Scripts Games Exercices Mail Version Franšaise Previous topic Next topic