3. Triangle $A B C$ is an isosceles right triangle: the angle at $C$ is right and $|A C|=|B C|=12$. Point $M$ is the midpoint of side $A B$. Point $D$ lies on side $A C$. Finally, point $E$ is the intersection of line segments $C M$ and $B D$, see the figure below.

If $|C D|=3$, what is the area of quadrilateral $A M E D$?

![](https://cdn.mathpix.com/cropped/2024_04_17_01cb006263f00c7722c1g-3.jpg?height=371&width=371&top_left_y=885&top_left_x=1522)
B3. $\frac{162}{5}$ The area of $\triangle A M C$ is half the area of $\triangle A B C$. The area of $\triangle A B C$ is $\frac{1}{2} \cdot|A C| \cdot|B C|=\frac{1}{2} \cdot 12 \cdot 12=72$. Therefore, the area of $\triangle A M C$ is 36. To determine the area of quadrilateral $A M E D$, we now only need to determine the area of $\triangle C D E$.

Let $F$ be the point obtained by reflecting $D$ over $C E$. Due to symmetry, we see that $\triangle C E F$ has the same area as $\triangle C D E$. Moreover, we also know that $|C F|=3$ and $|B C|=12$. Triangles $B C E$ and $C E F$

![](https://cdn.mathpix.com/cropped/2024_04_17_fc21619166b069277c26g-5.jpg?height=382&width=371&top_left_y=203&top_left_x=1528)
have the same height, but the base of $\triangle B C E$ is four times as large. Therefore, the area of $\triangle B C E$ is four times the area of $\triangle C E F$ (or $\triangle C D E$).

If we add triangle $C D E$ to this, we see that $\triangle B C D$ has five times the area of $\triangle C D E$. The area of $\triangle B C D$ is $\frac{1}{2} \cdot|C D| \cdot|B C|=\frac{1}{2} \cdot 3 \cdot 12=18$. Altogether, we find that $\triangle C D E$ has an area of $\frac{18}{5}$, and the area of quadrilateral $A M E D$ is therefore $36-\frac{18}{5}=\frac{162}{5}$.