The convex quadrilateral $ABCD$ has area $1$, and $AB$ is produced to $E$, $BC$ to $F$, $CD$ to $G$ and $DA$ to $H$, such that $AB=BE$, $BC=CF$, $CD=DG$ and $DA=AH$. Find the area of the quadrilateral $EFGH$.
1. **Cheap Method:**
   - Assume without loss of generality that $ABCD$ is a square with area $1$.
   - Since $AB = BE$, $BC = CF$, $CD = DG$, and $DA = AH$, each of the triangles $\triangle AHE$, $\triangle BEF$, $\triangle CFG$, and $\triangle DGH$ will have the same area as the square $ABCD$.
   - Therefore, the area of each of these triangles is $1$.
   - The total area of quadrilateral $EFGH$ is the sum of the area of $ABCD$ and the areas of the four triangles:
     \[
     [EFGH] = [ABCD] + [AHE] + [BEF] + [CFG] + [DGH] = 1 + 1 + 1 + 1 + 1 = 5
     \]
   - Thus, the area of quadrilateral $EFGH$ is $\boxed{5}$.

2. **Standard Method:**
   - Consider the triangles formed by extending the sides of $ABCD$:
     - $\triangle BCD$ and $\triangle CFD$ have the same base $CD$ and the same height, so $[BCD] = [CFD]$.
     - Similarly, $\triangle CFD$ and $\triangle DFG$ have the same base $DF$ and the same height, so $[CFD] = [DFG]$.
     - Therefore, $[CFG] = 2[BCD]$.
   - By similar reasoning:
     - $[BEF] = 2[ABC]$
     - $[AHE] = 2[ABD]$
     - $[DGH] = 2[ACD]$
   - Summing these areas and adding the area of $ABCD$:
     \[
     [EFGH] = [ABCD] + 2[ABC] + 2[ABD] + 2[ACD] + 2[BCD]
     \]
   - Since $[ABC] + [ABD] + [ACD] + [BCD] = [ABCD] = 1$:
     \[
     [EFGH] = 1 + 2 \cdot 1 = 1 + 4 = 5
     \]
   - Thus, the area of quadrilateral $EFGH$ is $\boxed{5}$.