69th Belarusian Mathematical Olympiad

Denote by $A_1$ the intersection point of the lines $A{X}_1$ and $C{X}_2$, and by $C_1$ denote the intersection point of the lines $A{X}_2$ and $C{X}_1$. Let $B_1$ be the intersection point of the lines $B{Y}_1$ and $D{Y}_2$, and let $D_1$ be the intersection point of the lines $B{Y}_2$ and $D{Y}_1$. The lines $A_{1}A$ and $C_{1}C$ are the altitudes of the triangle $A_1{C}_1{X}_2$, hence $X_1{X}_2\perp A_1{C}_1$. Similarly, $Y_1{Y}_2\perp B_1{D}_1$. Thus it is enough to prove that the angle between the lines $A_1{C}_1$ and $B_1{D}_1$ equals to the angle between the lines $AC$ and $BD$. Further we use only oriented angles. Since $A{A}_1\perp A{C}_1$ and $C{C}_1\perp C{A}_1$, the quadrilateral $AC{A}_1{C}_1$ is cyclic, hence $$\angle (AC,A{C}_1)=\angle (A_{1}C,A_1{C}_1).$$ Adding $\angle (AC,A_{1}C)$ to both sides, we obtain $$\angle (AC,A{C}_1)+\angle (AC,A_{1}C)=\angle (AC,A_{1}C)+\angle (A_{1}C,A_1{C}_1)=\angle (AC,A_1{C}_1).$$ Since $\angle (AC,A{C}_1)=90^{\circ }-\angle (A{A}_1,AC)$ and $\angle (AC,A_{1}C)=90^{\circ }-\angle (C{C}_1,AC)$, we can write $\angle (AC,A_1{C}_1)=-\angle (A{A}_1,AC)-\angle (C{C}_1,AC)$. Further, $$\angle (A{A}_1,AC)=\angle (A{A}_1,AB)+\angle (AB,AC)=\angle (A{A}_1,AD)+\angle (AD,AC).$$ Using $\angle (A{A}_1,AB)+\angle (A{A}_1,AD)=0^{\circ }$ we get $2\angle (A{A}_1,AC)=\angle (AB,AC)+\angle (AD,AC)$. Similarly $2\angle (C{C}_1,AC)=\angle (CB,AC)+\angle (CD,AC)$. Therefore $$2\angle (AC,A_1{C}_1)=-\angle (AB,AC)-\angle (BC,AC)-\angle (CD,AC)-\angle (DA,AC).(1)$$ Similarly $$2\angle (BD,B_1{D}_1)=-\angle (AB,BD)-\angle (BC,BD)-\angle (CD,BD)-\angle (DA,BD).(2)$$ Subtracting (2) from (1) we obtain $\angle (AC,A_1{C}_1)-\angle (BD,B_1{D}_1)=2\angle (AC,BD)$. Finally $$\begin{array}{rl}\angle (B_1{D}_1,A_1{C}_1)& =\angle (B_1{D}_1,BD)+\angle (BD,AC)+\angle (AC,A_1{C}_1)=\\ & =2\angle (AC,BD)-\angle (AC,BD)=\angle (AC,BD).\end{array}$$