Consider an inversion of pole $P$. It turns the circles $C_1$ and $C_2$ into two parallel lines $c_1$ and $c_2$, and the line $ABC$ into a circle $(C)$, which passes through $P$ and is tangent to $c_1$ in $A'$ (the image of $A$) and which cuts $c_2$ in $B',C'$ (the images of $B,C$ respectively. In these conditions, it's really easy to show that $PA'$ is the internal or external bisector of $\angle B'PC'$ because it's obvious that $A'$ is the midpoint of the arc $B'C'$.

Hey Grobber, you are really an inversion freak! A simple homothety is sufficient to solve it By the way, this problem was proposed by BELGIUM!

Arne wrote: A simple homothety is sufficient to solve it I solved it with the "Raakomtrekshoek", although I got no idea how to translate that in English... can u help me a hand Ernie? Besides, I think this should not be a too advanced problem... it was asked as a BMO round 1 question!

Peter means the alternating segment theorem Well, by far the simplest solution is the one using a homothety anyway

Just to give the solution everybody has been talking of: Let $A$ be on $C_2$ and let $\displaystyle \{A'\}=t(A)\cap C_1$, where $t(A)$ is the tangent at $A$ to $C_2$. The homothety of center $P$ taking $C_2$ to $C_1$ maps $A$ into $A'$ and $t(A)$ into $t(A')$ (the tangent at $A'$ to $C_1$). $t(A)$ and $t(A')$ are parallel and intersect $C_1$ at $B,C$ and $A'$ respectively. This shows that $BA'=A'C$ and $A'$ is the midpoint of the arc $\t{BC}$. So $PA'$ is one of the angle bisectors of $\angle BPC$. Lines $PA'$ and $PA$ are the same and we are done.

Attachments:

I suggest the proof of the converse i.e.: Let $PBC$ be a triangle and let $(C)$ be a circle tangent to $BC$ at $A$. If $PA$ is the bisector of $\angle BAC$, then $(C)$ is tangent to the circumcircle of $PBC$ in $P$.

orl wrote: Two circles $C_{1}$ and $C_{2}$ are (externally or internally) tangent at a point $P$. The straight line $d$ is tangent at $A$ to one of the circles and cuts the other circle at the points $B$ and $C$. Prove that the straight line $PA$ is an interior or exterior bisector of the angle $\angle BPC$. Denote : the intersection $M$ between the line $BC$ and the common tangent in the point $P$ ; the reflection $D$ of the point $A$ w.r.t. the point $M$. Then the points $A$, $D$ are harmonical conjugate w.r.t. the pair of the points $B$, $C$ because $MA=MD$ and $MA^{2}=MP^{2}=MB\cdot MC$ (see the equivalence $A3$ of the harmonical division in the lemma from the my message http://www.mathlinks.ro/Forum/viewtopic.php?t=46146 ). But $MB=MA=MD\Longrightarrow$ the rays $[PA$ and $[PD$ are the bisectors of the angles between the lines $PB$ and $PC$.

could you just..