Problem 1653

1. Define the setup and notation:
Let $S_1$ and $S_2$ be two equal circles intersecting at points $X$ and $Y$. Let $\ell$ be a line intersecting $S_1$ at points $A$ and $C$, and $S_2$ at points $B$ and $D$ in the order $A, B, C, D$ on $\ell$. Define circles $\Gamma_1$ and $\Gamma_2$ such that both touch $S_1$ internally and $S_2$ externally, and both touch the line $\ell$. Additionally, $\Gamma_1$ and $\Gamma_2$ lie in different half-planes relative to $\ell$ and touch each other.

2. **Introduce the midpoint $M$ and radius $r$:**
Let $M$ be the midpoint of the segment $XY$. Since $S_1$ and $S_2$ are equal circles, the distance from $M$ to $X$ and $Y$ is the same, denoted as $r$. Thus, $MX = MY = r$.

3. State and prove the lemma:
Lemma: For any circle $\Gamma$ touching $S_1$ internally at point $P$ and $S_2$ externally at point $Q$, the points $M, P, Q$ are collinear, and the power of $M$ with respect to $\Gamma$ is $r^2$.

Proof of Lemma:
- The points $P, Q, M$ are the internal and external similicenters of the pairs of circles $(\Gamma, S_1)$, $(\Gamma, S_2)$, and $(S_1, S_2)$ respectively.
- By Monge's Theorem, the three similicenters of any three circles are collinear. Therefore, $M, P, Q$ are collinear.
- Let $Q'$ be the reflection of $Q$ in $M$. Then, the power of $M$ with respect to $\Gamma$ is given by:
$$\text{Pow}_{\Gamma}(M) = MP \cdot MQ = MP \cdot MQ' = \text{Pow}_{S_1}(M) = r^2$$

4. **Apply the lemma to $\Gamma_1$ and $\Gamma_2$:**
By the lemma, $M$ has the same power with respect to both $\Gamma_1$ and $\Gamma_2$. Since $\Gamma_1$ and $\Gamma_2$ touch each other, their radical axis is the line $\ell$. Therefore, $M$ lies on $\ell$.

5. Conclude the symmetry argument:
Since $M$ lies on $\ell$ and is the midpoint of $XY$, the configuration is symmetric with respect to $M$. This symmetry implies that the segments $AB$ and $CD$ are equal in length.

$$\boxed{AB = CD}$$