12.9 Let $O$ be the circumcenter of $\triangle ABC$, and $K$ be the center of another circle, the two circles being tangent at $T$. $TE$ is the common tangent. Since $AB$, $AC$ are tangent to $\odot K$, and $AB=AC$, it can be deduced that $O$, $K$, $T$ all lie on the angle bisector of $\angle A$. $O K \cap P Q=I$, which is clearly the midpoint of $P Q$. $O I \perp P Q$, $\angle B T P=\angle E T P-\angle E T B=\angle B P T-\angle B A T=\angle P T A$, and $B$, $T$, $P$, $I$ are concyclic, $\angle P B I=\angle P T I=\angle B T P=\angle B I P$, it is easy to see that $P Q \parallel B C$, so $\angle P B I=\angle B I P=\angle I B C$, that is, $I B$ bisects $\angle A B C$, $I A$ bisects $\angle B A C$, thus, $I$ is the incenter.

For this problem, (1) the condition $AB=AC$ is redundant (2) if “$\odot K$ is internally tangent to $\odot O$” is changed to “$\odot K'$ is externally tangent to $\odot O$”, then the midpoint of segment $P Q$ is the excenter of $\triangle ABC$; (3) the internal common tangent of $\odot K$ and $\odot K'$ is tangent to $BC$. Refer to Shan Huan's "Methods for Solving International Mathematical Competition Problems", where (3) is a discovery by Ye Zhonghao.