Let I be the incenter of triangle ABC.

Prove that the intersection (not C) of line CI and the circumcircle of triangle ABC is the circumcenter of triangle ABI.

Hi!

Let be P the intersection between CI and the circumcircle of the ABC triangle.

In APBC we have :

<ACP=<PBA ; <BCP = <PAB

But because <ACP=<BCP we have <PBA=<PAB and PA=PB (1)

After that <AIP=<IAC+<ICA = (<A+<C)/2

              < PAI=<PAB+<BAI = (<A+<C)/2

So <AIP = <PAI and so PI=PA (2)

From (1) and (2) PA=PB=PI and p is the circumcenter of the ABI triangle.

Posted by Chesca Ciprian on 2009-01-28 16:57:44