Note that angles $B C E$ and $C B E$ are acute (as halves of the interior angles of a convex polygon). Since the angles at the common vertex $C$ of triangles $C D E$ and $C B E$ are equal, and $\angle C D E > \angle C B E$ (one is obtuse, the other is acute), then $\angle C E D < \angle C E B$. Therefore, if from the ray $E C$ in the half-plane containing point $B$, a ray is laid off at an angle equal to $\angle C E D$, the laid-off ray will lie between the sides of angle $C E B$, and thus will intersect segment $B C$ at some point $M$.

Triangles $C M E$ and $C D E$ are equal by the common side $E M$ and the two adjacent angles, and since $\angle B M E = 180^{\circ} - \angle C M E = 180^{\circ} - 145^{\circ} = 35^{\circ} = \angle B A E$, triangles $B M E$ and $B A E$ are also equal. Therefore, $S_{A B C D E} = 2 S_{B C E} = 22$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-48.jpg?height=375&width=509&top_left_y=1659&top_left_x=777)

## Answer

22.
3- |

In the convex pentagon $A B C D E$, the diagonals $B E$ and $C E$ are the angle bisectors of the angles at vertices $B$ and $C$ respectively, $\angle A=35^{\circ}, \angle D=145^{\circ}$, and the area of triangle $B C E$ is 11. Find the area of the pentagon $A B C D E$.