Solution. Since $S_{A B E}=S_{A B C}$, by the lemma proved above, we have $E C \| A B$. The other diagonals are also parallel to the corresponding sides. Let $P$ be the intersection point of $B D$ and $E C$. If $S_{B P C}=x$, then $S_{A B C D E}=$ $S_{A B E}+S_{E P B}+S_{E D C}+S_{B P C}=3+x\left(S_{E P B}=S_{A B E}=1\right.$, since $A B P E$ is a parallelogram). Note that $S_{B P C}=S_{D P E}=x$, since $B C D E$ is a trapezoid. Since $S_{B P C}: S_{D P C}=B P: D P=S_{E P B}: S_{E P D}$, then $x:(1-x)=1: x$, which means $x=(\sqrt{5}-1) / 2$ and $S_{A B C D E}=(\sqrt{5}+5) / 2$.

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4. Each diagonal of the convex pentagon $A B C D E$ cuts off a triangle of unit area from it. Calculate the area of the pentagon $A B C D E$.

Lemma. Let in a convex quadrilateral $A B C D$ the diagonals $A C$ and $B D$ intersect at point $O$. If triangles $A B O$ and $C O D$ are equal in area, then $A B C D$ is a trapezoid.

Proof. Since $S_{A O B}=S_{C O D}$, then $S_{A B D}=S_{C A D}$, which means the heights from points $B$ and $C$ to line $A D$ in triangles $A B D$ and $C A D$ are equal. This implies that all points on line $B C$ are equidistant from line $A D$, hence $B C \| A D$.