4. On the side $A D$ of the parallelogram $A B C D$, a point $E$ is taken such that $3 \cdot \overline{A E}=\overline{A D}$.

Find the ratios $\overline{A P}: \overline{P C}$ and $\overline{B P}: \overline{P E}$, where $P=A C \cap B E$.
Solution. Vectors $\overrightarrow{A P}$ and $\overrightarrow{A C}$ are collinear, so there exists a real number $x$ such that $\overrightarrow{A P}=x \overrightarrow{A C}$. Similarly, there exists a real number $y$ such that

$$
\overrightarrow{P E}=y \overrightarrow{B E}
$$

From the diagram, it is clear that

$$
\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{A D}, \overrightarrow{B E}=-\overrightarrow{A B}+\frac{1}{3} \overrightarrow{A D}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_c6c83167c56ccaf27c0cg-3.jpg?height=312&width=613&top_left_y=995&top_left_x=847)
thus we have

$$
\overrightarrow{A P}=x(\overrightarrow{A B}+\overrightarrow{A D}), \overrightarrow{P E}=y\left(-\overrightarrow{A B}+\frac{1}{3} \overrightarrow{A D}\right)
$$

Since $\overrightarrow{A P}+\overrightarrow{P E}+\overrightarrow{E A}=\overrightarrow{0}$, we will have

$$
x(\overrightarrow{A B}+\overrightarrow{A D})+y\left(-\overrightarrow{A B}+\frac{1}{3} \overrightarrow{A D}\right)-\frac{1}{3} \overrightarrow{A D}=\overrightarrow{0} \quad \Rightarrow \quad(x-y) \overrightarrow{A B}+\left(x+\frac{y}{3}-\frac{1}{3}\right) \overrightarrow{A D}=\overrightarrow{0}
$$

Vectors $\overrightarrow{A B}$ and $\overrightarrow{A D}$ are not collinear, so it follows that

$$
\left\{\begin{array}{l}
x-y=0 \\
x+\frac{y}{3}=\frac{1}{3}
\end{array}\right.
$$

from which we get $x=y=\frac{1}{4}$. Therefore, we have

$$
\overrightarrow{A P}=\frac{1}{4} \overrightarrow{A C}, \overrightarrow{P E}=\frac{1}{4} \overrightarrow{B E}
$$

and from this it follows that

$$
\overline{A P}: \overline{P C}=1: 3, \overline{B P}: \overline{P E}=3: 1
$$

## III year