14. Let points $E$, $F$, $G$, $H$ be on the sides $AB$, $BC$, $CD$, $DA$ of a quadrilateral $ABCD$ with area 1, respectively, and
$$
\frac{A E}{E B}=\frac{B F}{F C}=\frac{C G}{G D}=\frac{D H}{H A}=k \text { ( } k \text { is a positive number). }
$$

Find the area of quadrilateral $E F G H$.
14. As shown in Figure 7, connect $A C$. Given $\frac{C G}{G D}=\frac{D H}{H A}=k$, then
$$
\begin{array}{l}
\frac{D H}{D A}=\frac{k}{k+1}, \\
\frac{D P}{D A}=\frac{D G}{D C}=\frac{1}{k+1} .
\end{array}
$$

Therefore, $S_{\triangle D H C}=\frac{k}{(k+1)^{2}} \cdot S_{\triangle D A C}$.
Similarly, $S_{\triangle B E F}=\frac{k}{(k+1)^{2}} \cdot S_{\triangle B C C}$.
Thus, $S_{\triangle D H G}+S_{\triangle B E F}=\frac{k}{(k+1)^{2}}\left(S_{\triangle A C}+S_{\triangle B C C}\right)$
$$
=\frac{k}{(k+1)^{2}} \cdot S_{\text {quadrilateral } B C D D}=\frac{k}{(k+1)^{2}}.
$$

Connect $B D$. Similarly, $S_{\triangle B H}+S_{\triangle C F C}=\frac{k}{(k+1)^{2}}$. Therefore,
$$
\begin{array}{l}
S_{\text {quadrilateral } E F C H}=S_{\text {quadrilateral } B B C D}-\left(S_{\triangle A E H}+S_{\triangle B E F}+S_{\triangle C F C}+S_{\triangle H G C}\right) \\
=1-\frac{2 k}{(k+1)^{2}}=\frac{k^{2}+1}{(k+1)^{2}} .
\end{array}
$$