4.5. On the extensions of sides $D A, A B, B C$, $C D$ of a convex quadrilateral $A B C D$, points $A_{1}, B_{1}, C_{1}, D_{1}$ are taken such that $\overrightarrow{D A_{1}}=2 \overrightarrow{D A}$, $\overrightarrow{A B_{1}}=2 \overrightarrow{A B}, \overrightarrow{B C_{1}}=2 \overrightarrow{B C}$ and $\overrightarrow{C D_{1}}=2 \overrightarrow{C D}$. Find the area of the resulting quadrilateral $A_{1} B_{1} C_{1} D_{1}$, given that the area of quadrilateral $A B C D$ is $S$.
4.5. Since $A B=B B_{1}$, then $S_{B B_{1} C}=S_{B A C}$. And since $B C=C C_{1}$, then $S_{B_{1} C_{1} C}=S_{B B_{1} C}=S_{B A C}$ and $S_{B B_{1} C_{1}}=2 S_{B A C}$. Similarly $S_{D D_{1} A_{1}}=2 S_{A C D}$,
therefore $S_{B B_{1} C_{1}}+S_{D D_{1} A_{1}}=2 S_{A B C}+2 S_{A C D}=2 S_{A B C D}$. Similarly $S_{A A_{1} B_{1}}+$ $+S_{C C_{1} D_{1}}=2 S_{A B C D}$, therefore $S_{A_{1} B_{1} C_{1} D_{1}}=S_{A B C D}+S_{A A_{1} B_{1}}+S_{B B_{1} C_{1}}+S_{C C_{1} D_{1}}+$ $+S_{D D_{1} A_{1}}=5 S_{A B C D}$