CWMO-2001-6

[Tahmid Hasan]

$P$ is a point on the exterior of a circle centered at $O$.The tangents to the circle from $P$ touch the circle at $A$ and $B$.Let $Q$ be the point of intersection of $PO$ and $AB$.Let $CD$ be any chord of the circle passing through $Q$.Prove that $\triangle PAB$ and $\triangle PCD$ have the same incentre.

[Tahmid Hasan]

It looks as if this is the $2000^{th}$ topic
Anyway here's the killer lemma:

Let $\ell$ be the internal angle bisector of $\angle A$ in a $\triangle ABC$.
Let $D=\odot ABC \cap \ell$.$I$ be an arbitrary point on segment $AD$.Then $I$ is the incentre of $\triangle ABC$
iff $DB=DB=DI$.

[Phlembac Adib Hasan]

CWMO 2001-6.png (15.51KiB)Viewed 2565 times

Proof:
$PA=PB$,$A,O,B,P$ concyclic and $AO=OI=OB$.
So $I$ is the incenter of $\triangle ABP$.
$C,O,D,P$ con-cyclic, because $CQ\cdot QD=AQ\cdot QB=OQ\cdot QP$.(Using circle $AOBP$)
So $\angle CPO=\angle CDO=\angle OCD=\angle OPD$.Also $OC=OI=OD$ and so $I$ is also incenter of $\triangle PCD$.