(Solution 1): Let the point of tangency of $O$, $O_1$ be $P$, and the point of tangency of $O$, $O_2$ be $Q$; the original problem is equivalent to $PMQA'$ being a harmonic quadrilateral. Consider the transformation "inversion centered at $A$ with power $\overrightarrow{AB} \times \overrightarrow{AC}$, followed by reflection across the angle bisector of $\angle BAC$"; then $A$, $B$, $C$ remain fixed, $O_1$ is transformed into the $A$-excircle (let its center be $I_A$), and $O_2$ is transformed into the incircle (let its center be $I$). Let $D$, $Y$, $Z$ be the feet of the perpendiculars from $A$, $I$, $I_A$ respectively to $\overrightarrow{BC}$, and let $L$ be the intersection of the angle bisector of $\angle BAC$ with $\overrightarrow{BC}$; then $P$ is transformed to $Z$, $Q$ is transformed to $Y$, $A'$ is transformed to $D$, and $M$ is transformed to $L$; the original problem is equivalent to $(D, L; Y, Z)$ being a harmonic range, which is again equivalent to $(A, L; I, I_A)$ being a harmonic range. Since $\overrightarrow{AI} : \overrightarrow{IL} = \overrightarrow{AB} : \overrightarrow{BL} = \overrightarrow{AI_A} : \overrightarrow{LI_A}$, the original proposition holds.

Let $MA'$ meet $AB$ at $M_1$, meet $AC$ at $M_2$, let $O_1$ be tangent to $AB$ at $N_1$, $O_1$ be tangent to $AC$ at $N_2$, $O_2$ be tangent to $AB$ at $W_1$, $O_2$ be tangent to $AC$ at $W_2$. By Mannheim's theorem, $N_1$, $I$, $N_2$ are collinear, $W_1I_AW_2$ are collinear, and $N_1N_2 \perp AI$, $W_1W_2 \perp AI_A$. By the Incenter-Excenter Lemma, $IM = I_AM$, hence $N_1M_1 = W_1M_1 = N_2M_2 = W_2M_2$, from which it follows that $MA'$ is indeed the radical axis of $O_1$, $O_2$. This completes the proof.

It suffices to prove that $MA'$ is the radical axis of $O_1$, $O_2$; then $X$ is the radical center of the three circles, and the original proposition is thereby proved.
Let $MA'$ meet $AB$ at $M_1$, meet $AC$ at $M_2$, let $O_1$ be tangent to $AB$ at $N_1$, $O_1$ be tangent to $AC$ at $N_2$, $O_2$ be tangent to $AB$ at $W_1$, $O_2$ be tangent to $AC$ at $W_2$. By Mannheim's theorem, $N_1$, $I$, $N_2$ are collinear, $W_1I_AW_2$ are collinear, and $N_1N_2 \perp AI$, $W_1W_2 \perp AI_A$. By the Incenter-Excenter Lemma, $IM = I_AM$, hence $N_1M_1 = W_1M_1 = N_2M_2 = W_2M_2$, from which it follows that $MA'$ is indeed the radical axis of $O_1$, $O_2$. This completes the proof.