37. In an isosceles triangle \(ABC (|AB| = |BC|)\), a perpendicular to \(AE\) is drawn through the endpoint \(E\) of the bisector \(AE\) until it intersects the extension of side \(AC\) at point \(F\) (with \(C\) between \(A\) and \(F\)). It is known that \(|AC| = 2m, |FC| = m/4\). Find the area of triangle \(ABC\).
37. If $\widehat{B A C}=\widehat{B C A}=2 \alpha$, then by the Law of Sines we find

$$
|A E|=\frac{2 m \sin 2 \alpha}{\sin 3 \alpha},|A F|=\frac{|A E|}{\cos \alpha}=\frac{2 m \sin 2 \alpha}{\sin 3 \alpha \cos \alpha}
$$

Thus, $\frac{9}{4} m=\frac{2 m \sin 2 \alpha}{\sin 3 \alpha \cos \alpha}$, from which $\cos 2 \alpha=\frac{7}{18}, S_{\triangle A B C}=$ $=m^{2} \operatorname{tg} 2 \alpha=\frac{5 m^{2} \sqrt{11}}{7}$.