Let D the other intersection point of the circumcircle with the bisector of angle C, E the diametral point of D. The distances DA = DI = DB (see Bisector.html ). It follows from the similar triangles FCI and BED,

CI/IF = ED/DB => CI*ID = CI*DB = ED*IF = r*(2R).

The product CI*ID is the power of I with respect to the circumcircle, hence is equal to R

R > 2r. The equality R=2r is valid only for the equilateral triangle.

In fact, if the relation is valid, then triangle CIF and the right-angled triangle with hypotenuse 2R and one side DI are similar. But last right-angled triangle is equal to EBD. Thus DB=DI and this implies that the incenter of triangle ABC coincides with I. It follows that the inradius of ABC is also equal to r as claimed.

Euler.html

Poncelet.html

Poristic.html

Produced with EucliDraw© |