6.1. On the sides $A C$ and $B C$ of triangle $A B C$, points $D$ and $E$ are taken, respectively. Segments $A E$ and $B D$ intersect at point $F$. Find the area of triangle $C D E$, if the areas of triangles $A B F, A D F$, and $B E F$ are $1, \frac{1}{2}$, and $\frac{1}{4}$, respectively.
Answer. $\frac{15}{56} \approx 0.27$.

Solution. Let $S_{1}=S_{A F D}=\frac{1}{2}, S_{2}=S_{A B F}=1, S_{3}=S_{B E F}=\frac{1}{4}, S_{4}=S_{D E F}, S_{5}=S_{C D E}$. Since the ratio of the areas of triangles with equal heights is equal to the ratio of the lengths of the bases, then

$$
\frac{S_{4}}{S_{3}}=\frac{F D}{F B}=\frac{S_{1}}{S_{2}}, \quad \frac{S}{S_{3}+S_{4}}=\frac{C E}{B E}=\frac{S+S_{1}+S_{4}}{S_{2}+S_{3}}
$$

From this, we find

$$
S_{4}=\frac{S_{1} S_{3}}{S_{2}}, \quad S=\frac{S_{1} S_{3}\left(S_{2}+S_{3}\right)\left(S_{1}+S_{2}\right)}{S_{2}\left(S_{2}^{2}-S_{1} S_{3}\right)}=\frac{15}{56}
$$