(JBMO 1998 - Problem 2), discussed in class

Let $A B C D E$ be a convex pentagon such that $A B=A E=C D=B C+D E=1$ and $\widehat{A B C}=\widehat{D E A}=90^{\circ}$. Calculate the area of $A B C D E$.
Let $P$ be the point on the ray $[D E)$ such that $D P=1$. Since $\widehat{A E P}=\widehat{A B C}=$ $90^{\circ}$, $A B=A E=1$ and $B C=1-D E=E P$, the triangles $A E P$ and $A B C$ are directly isometric. It follows that $A P=A C$ and that the polygons $A B C D E$ and $A C D P$ have the same area.

Furthermore, since $A P=A C$ and $C D=D P=1$, the line $(A D)$ is the axis of symmetry of $A C D P$, so the area of $A C D P$ is 2 times the area of the triangle $A D P$. Finally, the triangle $A D P$ has base $P D=1$ and height $A E=1$, so its area is $\frac{1}{2}$, and thus the area of $A C D P$ and $A B C D E$ is 1.

![](https://cdn.mathpix.com/cropped/2024_05_10_831e903be1b8a1180f82g-66.jpg?height=490&width=553&top_left_y=324&top_left_x=797)