[ Area of a Triangle. ]

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 $A B C D E$.

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Since $S_{\mathrm{ABE}}=S_{\mathrm{ABC}}$, then $E C \| A B$. The other diagonals are also parallel to the corresponding sides. Let $P$ be the intersection point of $B D$ and $E C$. If $S_{\mathrm{BPC}}=x$, then $S_{\mathrm{ABCDE}}=S_{\mathrm{ABE}}+S_{\mathrm{EPB}}+S_{\mathrm{EDC}}+S_{\mathrm{BPC}}=3+x\left(S_{\mathrm{EPB}}=\right.$ $S_{\mathrm{ABE}}=1$, since $A B P E$ is a parallelogram). Since $S_{\mathrm{BPC}}: S_{\mathrm{DPC}}=B P: D P=S_{\mathrm{EPB}}: S_{\mathrm{EPD}}$, then $x:(1-x)=1: x$, which means $x=(\sqrt{5}-1) / 2$ and $S_{\mathrm{ABCDE}}=(\sqrt{5}+5) / 2$.