Belarusian Mathematical Olympiad

Denote by $A_1$ the intersection point of the lines $AX_1$ and $CX_2$, and by $C_1$ denote the intersection point of the lines $AX_2$ and $CX_1$. Let $B_1$ be the intersection point of the lines $BY_1$ and $DY_2$, and let $D_1$ be the intersection point of the lines $BY_2$ and $DY_1$. The lines $A_1A$ and $C_1C$ are the altitudes of the triangle $A_1C_1X_2$, hence $X_1X_2 \perp A_1C_1$. Similarly, $Y_1Y_2 \perp B_1D_1$. Thus it is enough to prove that the angle between the lines $A_1C_1$ and $B_1D_1$ equals to the angle between the lines $AC$ and $BD$. Further we use only oriented angles.
Since $AA_1 \perp AC_1$ and $CC_1 \perp CA_1$, the quadrilateral $ACA_1C_1$ is cyclic, hence
$$\angle(AC, AC_1) = \angle(A_1C, A_1C_1).$$
Adding $\angle(AC, A_1C)$ to both sides, we obtain
$$\angle(AC, AC_1) + \angle(AC, A_1C) = \angle(AC, A_1C) + \angle(A_1C, A_1C_1) = \angle(AC, A_1C_1).$$
Since $\angle(AC, AC_1) = 90^\circ - \angle(AA_1, AC)$ and $\angle(AC, A_1C) = 90^\circ - \angle(CC_1, AC)$, we can write $\angle(AC, A_1C_1) = -\angle(AA_1, AC) - \angle(CC_1, AC)$. Further,
$$\angle(AA_1, AC) = \angle(AA_1, AB) + \angle(AB, AC) = \angle(AA_1, AD) + \angle(AD, AC).$$
Using $\angle(AA_1, AB) + \angle(AA_1, AD) = 0^\circ$ we get $2\angle(AA_1, AC) = \angle(AB, AC) + \angle(AD, AC)$. Similarly $2\angle(CC_1, AC) = \angle(CB, AC) + \angle(CD, AC)$. Therefore
$$2\angle(AC, A_1C_1) = -\angle(AB, AC) - \angle(BC, AC) - \angle(CD, AC) - \angle(DA, AC). \quad (1)$$
Similarly
$$2\angle(BD, B_1D_1) = -\angle(AB, BD) - \angle(BC, BD) - \angle(CD, BD) - \angle(DA, BD). \quad (2)$$
Subtracting (2) from (1) we obtain $\angle(AC, A_1C_1) - \angle(BD, B_1D_1) = 2\angle(AC, BD)$.
Finally
$$\begin{aligned} \angle(B_1D_1, A_1C_1) &= \angle(B_1D_1, BD) + \angle(BD, AC) + \angle(AC, A_1C_1) = \\ &= 2\angle(AC, BD) - \angle(AC, BD) = \angle(AC, BD). \end{aligned}$$