In an acute-angled triangle $ABC$ on the side $AC$, point $P$ is chosen in such a way that $2AP = BC$. Points $X$ and $Y$ are symmetric to $P$ with respect to vertices $A$ and $C$, respectively. It turned out that $BX = BY$.  Find $\angle BCA$.
1. **Assign coordinates to points:**
   Let $A(0,0)$, $B(a,b)$, and $C(c,0)$. The length of $BC$ is given by:
   \[
   BC = \sqrt{(a-c)^2 + b^2} = w
   \]
   Given that $2AP = BC$, we have $AP = \frac{w}{2}$. Since $P$ is on $AC$, we can place $P$ at $(\frac{w}{2}, 0)$.

2. **Find coordinates of $X$ and $Y$:**
   - $X$ is symmetric to $P$ with respect to $A$, so $X$ has coordinates $(-\frac{w}{2}, 0)$.
   - $Y$ is symmetric to $P$ with respect to $C$, so $Y$ has coordinates $(2c - \frac{w}{2}, 0)$.

3. **Use the condition $BX = BY$:**
   The condition $BX = BY$ implies:
   \[
   BX^2 = BY^2
   \]
   Calculate $BX^2$:
   \[
   BX^2 = \left(a + \frac{w}{2}\right)^2 + b^2
   \]
   Calculate $BY^2$:
   \[
   BY^2 = \left(a - (2c - \frac{w}{2})\right)^2 + b^2 = \left(a - 2c + \frac{w}{2}\right)^2 + b^2
   \]
   Set these equal:
   \[
   \left(a + \frac{w}{2}\right)^2 + b^2 = \left(a - 2c + \frac{w}{2}\right)^2 + b^2
   \]
   Simplify:
   \[
   \left(a + \frac{w}{2}\right)^2 = \left(a - 2c + \frac{w}{2}\right)^2
   \]
   Expand both sides:
   \[
   a^2 + a w + \frac{w^2}{4} = a^2 - 4ac + 4c^2 + a w - 2cw + \frac{w^2}{4}
   \]
   Cancel out common terms:
   \[
   0 = -4ac + 4c^2 - 2cw
   \]
   Factor out common terms:
   \[
   0 = 2c(2c - 2a - w)
   \]
   Since $c \neq 0$ (as $C$ is not at the origin), we have:
   \[
   2c - 2a - w = 0 \implies w = 2(c - a)
   \]

4. **Find $b$ using $w$:**
   Substitute $w = 2(c - a)$ into the expression for $w$:
   \[
   w = \sqrt{(a - c)^2 + b^2} = 2(c - a)
   \]
   Square both sides:
   \[
   (a - c)^2 + b^2 = 4(c - a)^2
   \]
   Simplify:
   \[
   (a - c)^2 + b^2 = 4(a - c)^2
   \]
   Subtract $(a - c)^2$ from both sides:
   \[
   b^2 = 3(a - c)^2
   \]
   Take the square root:
   \[
   b = \sqrt{3}(a - c)
   \]

5. **Find $\tan \angle BCA$:**
   The tangent of $\angle BCA$ is given by:
   \[
   \tan \angle BCA = \frac{b}{a - c} = \frac{\sqrt{3}(a - c)}{a - c} = \sqrt{3}
   \]
   Since $\angle BCA$ is in an acute-angled triangle, we have:
   \[
   \angle BCA = 60^\circ
   \]

The final answer is $\boxed{60^\circ}$