3. Given a convex quadrilateral $ABCD$ with area $P$. We extend side $AB$ beyond $B$ to $A_1$ such that $\overline{AB}=\overline{BA_1}$, then $BC$ beyond $C$ to $B_1$ such that $\overline{BC}=\overline{CB_1}$, then $CD$ beyond $D$ to $C_1$ so that $\overline{CD}=\overline{DC_1}$, and $DA$ beyond $A$ to $D_1$ such that $\overline{DA}=\overline{AD_1}$. What is the area of quadrilateral $A_1B_1C_1D_1$?
Solution. We connect $A$ with $C_{1}$, $B$ with $D_{1}$, $C$ with $A_{1}$, and $D$ with $B_{1}$. Then,

$$
\begin{aligned}
& P_{\triangle A B C}=P_{I}=P_{I I} \\
& P_{\triangle B C D}=P_{I I I}=P_{I V} \\
& P_{\triangle C D A}=P_{V}=P_{V I} \\
& P_{\triangle D A B}=P_{V I I}=P_{I I I}
\end{aligned}
$$

where

$$
\begin{array}{ll}
P_{I}=P_{\Delta A_{1} C}, & P_{I I}=P_{\Delta C A_{1} B_{1}} \\
P_{I I I}=P_{\Delta D C B_{1}}, & P_{I V}=P_{\Delta C_{1} D B_{1}} \\
P_{V}=P_{\Delta C_{1} D A}, & P_{V I}=P_{\Delta C_{1} D_{1} A} \\
P_{I I I}=P_{\Delta A D_{1} B}, & P_{V I I I}=P_{\Delta B D_{1} A_{1}}
\end{array}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_ea01241e897ed9d0948dg-02.jpg?height=507&width=641&top_left_y=708&top_left_x=838)

We have:

$$
\begin{aligned}
P_{\Delta A B C}+P_{\Delta B C D}+P_{\Delta C D A}+P_{\Delta D A B} & =P_{I}+P_{I I I}+P_{V}+P_{I I}= \\
& =P_{I I}+P_{I V}+P_{V I}+P_{V I I I}=2 P \\
P_{1}=P+\left(P_{I}+P_{I I}+P_{I I I}+P_{I V}+P_{V}+P_{I I}\right. & \left.+P_{V I I}+P_{I I I}\right)=P+(2 P+2 P)=5 P
\end{aligned}
$$