Let B be the other intersection of the circles $C_1, C_2$ and $H \equiv P_1P_2 \cap Q_2Q_2$ the intersection of their common external tangents, identical with their external homothety center lying on the center line $O_1O_2$. Since the points pairs $(P_1, P_2), (Q_1Q_2)$ are centrally similar with the homothety center H and coefficient $\frac{r_1}{r_2}$, where $r_1, r_2$ are radii of the circles $C_1, C_2$, the midpoints $(M_1, M_2)$ are centrally similar with the same homothety center and coefficient. Let the line HA meet the circles $C_1, C_2$ at points $R_1, R_2$, respectively, different from the point A. Again, the points pairs $(R_1, A),\ (A, R_2)$ are centrally similar with the same homothety center and coefficient. Therefore, the triangles $\triangle M_1R_1A \sim \triangle M_2AR_2$ are centrally similar with the same homothety center and coefficient. Thus their corresponding sides $M_1R_1 \parallel M_2A,\ M_1A \parallel M_2R_2$ are parallel. Let B' be the intersection of the lines $M_1R_1, M_2R_2$, forming a parallelogram $M_1AM_2B'$. The radical axis AB of the circles $C_1, C_2$ cuts the tangent length segments $P_1P_2, Q_1Q_2$ at their midpoints, hence, it also cuts the segment $M_1M_2$ at its midpoint N. As $M_1M_2$ is one tdiagonal of the parallelogram $M_1AM_2B'$, the line AB is the other diagonal line of this parallelogram. Since AN = BN, the vertex B' of this parallelogram is identical with the point B and the parallelogram is a rhombus (perpendicular diagonals or all sides equal). As a result, $\angle M_1AM_2 = \angle M_1BM_1 \equiv \angle R_1BR_2 = 180^\circ - (\angle AR_1B + \angle AR_2B) =$ $= 180^\circ - \frac{\angle AO_1B + \angle AO_2B}{2} = 180^\circ - (\angle O_2O_1A + \angle O_1O_2A) = \angle O_1AO_2$

Little more general: Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. Let $S$ be such point on line $O_1O_2$ so that tangents on $C_1$ touches it at $P_1$ and $Q_1$ and tangents on $C_2$ touches it at $P_2$ and $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2 = \angle M_1AM_2$. Proof: Since $S$ is image of $M_1$ under inversion wrt circle $C_1$ we have: \[ \angle O_1AM_1 = \angle O_1M_1'A'= \angle O_1SA \] Since $S$ is image of $M_2$ under inversion wrt circle $C_2$ we have:\[ \angle O_2SA= \angle O_2A'S'= \angle O_2AM_2 \] Image of $A$ is in both cases $A$ itself, since it lies on both circles. Since $\angle O_1SA = \angle O_2SA$ we have: \[ \angle M_1AO_1=\angle M_2AO_2 \] Now: \[ \angle O_1AO_2 = \angle M_1AM_2-\angle M_1AO_1+\angle M_2AO_2 = \angle M_1AM_2 \]

orl wrote: Let $ A$ be one of the two distinct points of intersection of two unequal coplanar circles $ C_1$ and $ C_2$ with centers $ O_1$ and $ O_2$ respectively. One of the common tangents to the circles touches $ C_1$ at $ P_1$ and $ C_2$ at $ P_2$, while the other touches $ C_1$ at $ Q_1$ and $ C_2$ at $ Q_2$. Let $ M_1$ be the midpoint of $ P_1Q_1$ and $ M_2$ the midpoint of $ P_2Q_2$. Prove that $ \angle O_1AO_2 = \angle M_1AM_2$. I think this has a similar idea as yetti's solution, but a bit different:

Let $B$ be the other intersection point of these two circles. Let $P_1P_2$, $Q_1Q_2$, $O_1O_2$ meet at $Q$. Let $AB$ meet $P_1P_2$ at $P$. Clearly, $P_1Q_1$ and $O_1O_2$ are perpendicular at $M_1$; $P_2Q_2$ and $O_1O_2$ are perpendicular at $M_2$. Since $\triangle O_1P_1M_1 \sim \triangle O_2P_2M_2$, \[\dfrac{O_1P_1}{O_1M_1} = \dfrac{O_2P_2}{O_2M_2} \Rightarrow \dfrac{O_1A}{O_1M_1} = \dfrac{O_2A}{O_2M_2} \qquad{(*)}\] Since $P$ is on the radical axis, $PP_1 = PP_2$, so in the right trapezoid $P_1P_2M_2M_1$, $AB$ is the midsegment. So we have $M_1A = AM_2$ and $\angle AM_1O_1 = \angle AM_2O_1$. Let $M$ be a point on $O_1O_2$ such that $\triangle AM_1O_1 \cong \triangle AM_2M$ (which means $AM=AO_1$ ve $MM_2 = M_1O_1$). So, from $(*)$, we get \[\dfrac{AM}{MM_2} = \dfrac{O_2A}{O_2M_2}\] This means, $AM_2$ is the angle bisector of $\angle MAO_2$. So, \[\angle O_2AM_2 = \angle M_2AM= \angle M_1AO_1 \Longrightarrow \angle M_1AM_2 = \angle O_1AO_2.\]

It is similar to all but slightly shorter...Let, $P_1P_2,Q_1Q_2,O_1O_2$ concur at $Q$.Then a homothety with centre $Q$ that sends $C_1$ to $C_2$.Let $QA \cap C_1 =B$.Under the homothety $A$ is the image of $B$.So, $\angle M_1BO_1 =\angle M_2AO_2$ and $\triangle QP_1O_1 \sim \triangle QM_1P_1 \implies \frac{QO_{1}}{QP_{1}} = \frac{QP_{1}}{QM_{1}}$.so $QO_1.QM_1 = Q.P^2_1 = QA .QB$ Hence $A,M_1,B,O_1$ are concyclic.so $\angle O_1BM_1=\angle O_1AM_1 \implies \angle O_1AM_1= \angle O_2AM_2 \implies \angle O_1AO_2=\angle M_1AM_2 $

I'm sorry that I can't use LaTeX because I am new user and don't have permission to add images to posts. It can be easily solved with this lemma: Let A-symmedian of triangle ABC intersect its circumscribed circle u at D, P on AD satisfying AP = PD. Let us define center of u as O. Then B, C, P, O are concyclic. Now let S be intersection of tangents to circles from the problem, AS intersects C_1 at B. One can prove that AP_1BQ_1 is harmonic qudrilateral, so P_1Q_1 is symmedian of triangle AP_1B. By lemma we get that B, A, O_1, M_1 are concyclic, thus angle O_1AM_1 is equivalent to angle O_1BM_1. By homotety we get angle O_1BM_1 = angle O_2AM_2. Q. E. D.

To $\LaTeX$ this, It can be easily solved with this lemma: Let $A$-symmedian of $\triangle ABC$ intersect its circumscribed circle $u$ at $D, P$ on $AD$ satisfying $AP = PD$. Let us define center of $u$ as $O$. Then $(B, C, P, O)$ are concyclic. Now let $S$ be intersection of tangents to circles from the problem, $AS$ intersects $ C_1$ at $B$. One can prove that $\square AP_1BQ_1$ is harmonic qudrilateral, so $P_1Q_1$ is symmedian of $\triangle AP_1B$. By lemma we get that $(B, A, O_1, M_1)$ are concyclic, thus $\angle O_1AM_1$ is equivalent to $\angle O_1BM_1$. By homothety we get $\angle O_1BM_1 = \angle O_2AM_2$. $Q. E. D.$

Let $P$ be the exsimilicenter of $C_1$ and $C_2$, and let $\infty$ be the point at infinity on line $l=O_1O_2$. Then the pairs of points $(P, \infty)$, $(M_1, O_2)$ and $(O_1, M_2)$ are pairs under an involution on line $l$ (this involution is inversion on $P$). That means that $(AP, A\infty)$, $(AM_1, AO_2)$ and $(AO_1, AM_2)$ are also pairs under an involution on the pencil of lines through $A$. Since by inversion $$\angle PAM=\angle PO_2A=\angle O_2A\infty,$$this involution must be a reflection over the angle bisector of $\angle MAO_2$, which solves the problem. $\blacksquare$