, 2022

Proof. Denote the centers of the two circles by $O_1, O_2$, and the radii by $r_1, r_2$, respectively, where $r_1 > r_2$. We first show that there are two points $P, Q$ on the ray $O_1O_2$ such that $O_1P \cdot O_1Q = r_1^2$ and $O_2P \cdot O_2Q = r_2^2$.
One can choose a point $K$ on the ray $O_1O_2$ such that $O_1K = \frac{O_1O_2^2 + r_1^2 - r_2^2}{2O_1O_2}$. Since $O_1O_2 \le r_1 - r_2$, we have $O_1K \ge r_1$. We choose two points $P, Q$ on the line $O_1O_2$ such that $KP = KQ = \sqrt{O_1K^2 - r_1^2}$ so that $O_1P \cdot O_1Q = r_1^2$. On the other hand,
$$\begin{aligned} O_2P \cdot O_2Q - O_1P \cdot O_1Q &= O_2K^2 - O_1K^2 = (O_1K - O_1O_2)^2 - O_1K^2 \\ &= O_1O_2^2 - 2O_1O_2 \cdot O_1K = r_2^2 - r_1^2, \end{aligned}$$
which implies that $O_2P \cdot O_2Q = r_2^2$.
For any given line $\ell$ as in the problem, if it is perpendicular to $O_1O_2$, we certainly have $\angle APC = \angle DPB$ by symmetry. If not, from $O_2C^2 = O_2Q \cdot O_2P$ we know that $\triangle O_2CQ \sim \triangle O_2PC$. So $\frac{CQ}{CP} = \frac{r_2}{O_2P}$. Similarly, we obtain $\frac{DQ}{DP} = \frac{r_2}{O_2P}$, and therefore $\frac{CQ}{CP} = \frac{DQ}{DP}$, which means that the bisectors of $\angle CPD$ and $\angle CQD$ meet $\ell$ at the same point, say $M$. A similar argument shows that the bisectors of $\angle APB$ and $\angle AQB$ intersect $\ell$ at the same point again, say $M'$.

Note that the points $P, Q, M$ are all on an Apollonian circle with distance ratio $\frac{CQ}{DQ}$ to $C$ and $D$. The center of the Apollonian circle must be on $\ell$. Thus this center must be the intersection of $\ell$ with the perpendicular bisector of $PQ$. Similarly, the points $P, Q, M'$ are on the Apollonius circle with distance ratio $\frac{AQ}{BQ}$ to $A$ and $B$, whose center is the same point as above. Note that $K$ does not lie in the interior of $\Gamma_1$, so the center of this circle, denoted by $L$, must be outside $\Gamma_1$ so that $M$ coincides with $M'$. Therefore,
$$\angle APC = \angle APM - \angle CPM = \angle BPM - \angle DPM = \angle BPD.$$
This completes the proof.