Let $\omega_1$ and $\omega_2$ be two circles with centers $C_1$ and $C_2$, respectively, which intersect at two points $P$ and $Q$. Suppose that the circumcircle of triangle $PC_1C_2$ intersects $\omega_1$ at $A \neq P$ and $\omega_2$ at $B \neq P$. Suppose further that $Q$ is inside the triangle $PAB$. Show that $Q$ is the incenter of triangle $PAB$.
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Question [ China TST 2002 Quiz 6 P2] : $ \odot O_1$ and $ \odot O_2$ meet at points $ P$ and $ Q$. The circle through $ P$, $ O_1$ and $ O_2$ meets $ \odot O_1$ and $ \odot O_2$ at points $ A$ and $ B$. Prove that the distance from $ Q$ to the lines $ PA$, $ PB$ and $ AB$ are equal.
(Prove the following three cases: $ O_1$ and $ O_2$ are in the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ and $ O_2$ are out of the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ is in the common space of $ \odot O_1$ and $ \odot O_2$, $ O_2$ is out of the common space of $ \odot O_1$ and $ \odot O_2$
Solution: This one is quite easy compared to the other China TST problems. Note that $\triangle O_1PO_2\cong\triangle O_1QO_2\implies \angle O_2O_1P=\angle O_2O_1B.$ Also, $\angle PO_1O_2=\angle PAO_2,$ and $\angle QO_1O_2=\angle BAO_2,$ leading to $\angle PAQ=\angle BAQ.$ So $AQ$ bisects $\angle PAB.$ Similarly $BQ$ bisects $\angle PBA,$ so $Q$ is the incentre of $\triangle PAB.$ So it is equidistant from $PA,PB$ and $AB.\Box$
