12.007. In rectangle $A B C D$ ($A B \| C D$), triangle $A E F$ is inscribed. Point $E$ lies on side $B C$, point $F$ - on side $C D$. Find the tangent of angle $E A F$, if $A B: B C = B E: E C = C F: F D = k$.
Solution.

By the condition $\frac{A B}{B C}=\frac{B E}{E C}=\frac{C F}{F D}=k$ (Fig. 12.10). Let $A D=B C=x$, then $A B=D C=k x \cdot B C=B E+E C=k E C+E C$, hence $E C=\frac{B C}{1+k}=\frac{x}{1+k}$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-721.jpg?height=286&width=532&top_left_y=93&top_left_x=124)

Fig. 12.10

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-721.jpg?height=258&width=582&top_left_y=121&top_left_x=653)

Fig. 12.11

$B E=\frac{k x}{1+k} \cdot D C=F C+D F=k D F+D F$, then $D F=\frac{k x}{1+k}, F C=\frac{k^{2} x}{1+k}$. $\angle F A E=\angle D A E-\angle D A F, \angle D A E=\angle B E A \quad(B C \| A D, A E$ - transversal), hence $\angle F A E=\angle B E A-\angle D A F$. We have $\operatorname{tg} \angle F A E=$ $=\operatorname{tg}(\angle B E A-\angle D A F)=\frac{\operatorname{tg} \angle B E A-\operatorname{tg} \angle D A F}{1+\operatorname{tg} \angle B E A \cdot \operatorname{tg} \angle D A F}$. Consider $\triangle A D F$, $\angle A D F=90^{\circ}, \operatorname{tg} \angle D A F=\frac{D F}{A D}=\frac{k x}{(1+k) x}=\frac{k}{1+k} \cdot$ From $\triangle A B E, \angle A B E=90^{\circ}$, $\operatorname{tg} \angle B E A=\frac{A B}{B E}=\frac{k x(1+k)}{k x}=1+k$. Finally, $\operatorname{tg} \angle F A E=\frac{1+k-\frac{k}{1+k}}{1+\frac{k}{1+k} \cdot(1+k)}=$ $=\frac{k^{2}+k+1}{(1+k)^{2}}$.

Answer: $\frac{k^{2}+k+1}{(1+k)^{2}}$.