5. The figure shows a quadrilateral $A B C D$ in which $A D=D C$ and $\angle A D C=\angle A B C=90^{\circ}$. The point $E$ is the foot of the perpendicular from $D$ to $A B$. The length $D E$ is 25 . What is the area of quadrilateral $A B C D$ ?
5. 625

Draw a line through $D$ that is parallel to $A B$. Let $F$ be the intersection of that line with $B C$ extended, as shown in the diagram.

Now $\quad \angle E D C+\angle C D F=\angle E D C+\angle A D E=90^{\circ}$
Therefore $\angle C D F=\angle A D E=90^{\circ}-\angle E C D$
Also $\quad \angle D F C=\angle D E A=90^{\circ}$
Therefore triangles $A D E$ and $C D F$ are similar because they have the same set of angles. Because $A D=C D$ they must also be congruent.
Therefore the area of quadrilateral $A B C D$ is equal to the area of rectangle $F D E B$. By the congruent triangles $A D E$ and $C D F$ we know that $D F=D E=25$. Therefore the required area is $25 \times 25=625$.