SOLUTION. Let $a$ be the length of sides $A B$ and $C D$, and $v$ be the distance between their lines, which is also the height of triangle $A F D$ from vertex $A$ (Fig. 1). From the condition $E F \| B D$, according to theorem $u$, it follows that triangles $B C D$ and $E C F$ are similar; let $k \in(0,1)$ be the similarity coefficient. Once we calculate it, the problem will be solved.

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Fig. 1

Since $|F C|=k a,|F D|=(1-k) a$ and the heights of triangles $E C F, A B E$ from their common vertex $E$ have lengths $k v$, respectively $(1-k) v$, for the areas of triangles $A F D$ and $A B E$ we have

$$S_{A F D}=\frac{(1-k) a v}{2}=\frac{a(1-k) v}{2}=S_{A B E},$$

so both areas are equal for any $k \in(0,1)$. Since the area of triangle $E C F$ is $S_{E C F}=\frac{1}{2} k a \cdot k v=\frac{1}{2} k^{2} a v$ and the area of the entire parallelogram $A B C D$ is given by the formula $S_{A B C D}=a v$, we can express the area of triangle $A E F$ as follows:

$$\begin{aligned} S_{A E F} & =S_{A B C D}-S_{A B E}-S_{E C F}-S_{A F D}= \\ & =a v\left(1-\frac{1}{2}(1-k)-\frac{1}{2} k^{2}-\frac{1}{2}(1-k)\right)=a v\left(k-\frac{1}{2} k^{2}\right) . \end{aligned}$$

The areas of triangles $A B E, A F D$ will be equal to the area of triangle $A E F$ precisely when

$$\frac{1}{2}(1-k)=k-\frac{1}{2} k^{2} \quad \text { or } \quad k^{2}-3 k+1=0 \text {. }$$

This quadratic equation has roots

$$k_{1,2}=\frac{3 \pm \sqrt{5}}{2}$$

of which only the root $k=\frac{1}{2}(3-\sqrt{5})$ satisfies the condition $k \in(0,1)$. Let us add that for such $k$ we have

$$1-k=\frac{\sqrt{5}-1}{2}=\frac{k}{1-k}$$

Answer. The desired points $E, F$ are determined by the ratios

$$|C E|:|E B|=|C F|:|F D|=(\sqrt{5}-1): 2 \text {. }$$

Note. The equality $(1-k): 1=k:(1-k)$ from the conclusion of the solution means that points $E, F$ divide the corresponding sides of the parallelogram in the so-called golden ratio. This is expressed by the equalities

$$|C E|:|E B|=|E B|:|B C| \text { and } \quad|C F|:|F D|=|F D|:|D C| \text {. }$$

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