12.169. In the square \(ABCD\), an isosceles triangle \(AEF\) is inscribed; point \(E\) lies on side \(BC\), point \(F\) lies on side \(CD\), and \(AE = EF\). The tangent of angle \(AEF\) is 2. Find the tangent of angle \(FEC\).

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0923.jpg?height=466&width=470&top_left_y=78&top_left_x=96)

Fig. 12.39

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0923.jpg?height=458&width=656&top_left_y=82&top_left_x=609)

Fig. 12.40
## Solution.

Let $A B=a, \angle A E F=\alpha, \angle F E C=x$ (Fig. 12.39).

Then $\angle B E A=180^{\circ}-(\alpha+x)$.

In $\triangle A B E\left(\angle A B E=90^{\circ}\right)$ :

$$
\begin{aligned}
& A E=\frac{A B}{\sin \angle B E A}=\frac{a}{\sin (\alpha+x)} \\
& B E=A B \operatorname{ctg} \angle B E A=-a \operatorname{ctg}(\alpha+x)
\end{aligned}
$$

$$
E F=A E=\frac{a}{\sin (\alpha+x)}
$$

In $\triangle E C F\left(\angle E C F=90^{\circ}\right): E C=E F \cos \angle F E C=\frac{a \cos x}{\sin (\alpha+x)}$. $B E+E C=B C$.

Then $\frac{a \cos x}{\sin (\alpha+x)}-a \operatorname{ctg}(\alpha+x)=a \Leftrightarrow \cos x-\cos (\alpha+x)=\sin (\alpha+x) \Leftrightarrow \Leftrightarrow$ $\cos x-\cos \alpha \cos x+\sin \alpha \sin x=\sin \alpha \cos x+\cos \alpha \sin x \Leftrightarrow \frac{1}{\cos \alpha}-1+\operatorname{tg} \alpha \operatorname{tg} x=$ $=\operatorname{tg} \alpha+\operatorname{tg} x \Leftrightarrow \sqrt{1+\operatorname{tg}^{2} \alpha}-1+\operatorname{tg} \alpha \operatorname{tg} x=\operatorname{tg} \alpha+\operatorname{tg} x$. Since by the condition $\operatorname{tg} \alpha=2$, then $\sqrt{5}-1+2 \operatorname{tg} x=2+\operatorname{tg} x ; \operatorname{tg} x=3-\sqrt{5}$.

Answer: $3-\sqrt{5}$.