376. Given a triangle $A B C$, the angles of which are $\alpha, \beta$ and $\gamma$. Triangle $D E F$ is circumscribed around triangle $A B C$ such that vertices $A, B$ and $C$ are located on sides $E F$, $F D$ and $D E$ respectively, and $\angle E C A = \angle D B C = \angle F A B = \varphi$. Determine the value of the angle $\varphi$ for which the area of triangle $E F D$ reaches its maximum value.
376. Let's describe circles around triangles $A B F, B C D$ and $C A E$. They have a common point $M$. Since the angles of triangle $DEF$ are constant, $\angle D=\gamma, \angle E=\alpha, \angle F=\beta$, the constructed circles and point $M$ do not depend on $\varphi$. The side $D F$ (and therefore $E F$ and $E D$) will be the largest when $D F$ is perpendicular to $B M$. Let $\varphi_{0}$ be the angle corresponding to this position. Then $\angle M B C=\angle M C A=\angle M A B=90^{\circ}-\varphi_{0}$. Extend $C M$ to intersect the circumcircle of triangle $A M B$ at point $F_{1}$. It can be found that $\angle F_{1} B A=\alpha$, $\angle F_{1} A B=\beta ; F_{1} B$ turns out to be parallel to $A C$. Drop perpendiculars $F_{1} N$ and $B L$ from $F_{1}$ and $B$ to $A C$. Since $\left|F_{1} N\right|=|B L|$, then

$$
\operatorname{tg} \varphi_{0}=\operatorname{ctg}\left(90^{\circ}-\varphi_{0}\right)=\frac{|C N|}{\left|F_{1} N\right|}=\frac{|A N|}{\left|F_{1} N\right|}+\frac{|A L|}{|B L|}+\frac{|C L|}{|B L|}=\operatorname{ctg} \beta+
$$

$+\operatorname{ctg} \alpha+\operatorname{ctg} \gamma$. Thus, $\operatorname{tg} \varphi_{0}=\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma$. Note. The angle $\omega=90^{\circ}-\varphi_{0}$ is called the Brocard angle, and point $M$ is the Brocard point. For each triangle, there are two Brocard points. The position of the second point $M_{1}$ is determined by the condition $\angle M_{1} B A=\angle M_{1} A C=\angle M_{1} C B$.