Task B-1.5. Let $E$ be the midpoint of side $\overline{A B}$ of square $A B C D$ and let $P$ be the point of intersection of segment $\overline{D E}$ and diagonal $\overline{A C}$. Determine the area of triangle $A E P$ if $|A B|=2$ $\mathrm{cm}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_0148fcdd1c92efebb536g-03.jpg?height=794&width=745&top_left_y=535&top_left_x=663)

Let $P$ be the intersection of line segment $\overline{D E}$ and diagonal $\overline{A C}$. Let $v_{1}$ denote the height of triangle $A E P$ to the base $\overline{A E}$, and $v_{2}$ the height of triangle $C D P$ to the base $\overline{C D}$.

Triangles $A E P$ and $C D P$ are similar because they have the same angles.

(1 point)

Then,

$$
\frac{v_{1}}{v_{2}} = |A E| : |C D| = 1 : 2
$$

Also,

$$
2 = |A B| = |B C| = v_{1} + v_{2} = v_{1} + 2 v_{1} = 3 v_{1}
$$

so $v_{1} = \frac{2}{3} \text{ cm}$.

(1 point)

It follows that the area of triangle $A E P$ is

$$
P = \frac{1}{2} |A E| \cdot v_{1} = \frac{1}{2} \cdot \frac{1}{2} |A B| \cdot v_{1} = \frac{1}{4} \cdot 2 \cdot \frac{2}{3} = \frac{1}{3} \text{ cm}^2
$$