4. Let $E$ be the midpoint of side $A B$ of parallelogram $A B C D$ and let $M=A C \cap D E$. Find the ratios $\overline{A M}: \overline{M C}$ and $\overline{D M}: \overline{M E}$.
Solution. Let $\overrightarrow{A B}=\vec{a}$ and $\overrightarrow{A D}=\vec{b}$ (see the diagram). We have:

$$
\begin{aligned}
& \overrightarrow{A M}=\lambda \overrightarrow{A C}=\lambda(\vec{a}+\vec{b}) \\
& \overrightarrow{D M}=\mu \overrightarrow{D E}=\mu\left(-\vec{b}+\frac{1}{2} \vec{a}\right)
\end{aligned}
$$

for some numbers $\lambda$ and $\mu$. Since

$$
\overrightarrow{A D}+\overrightarrow{D M}=\overrightarrow{A M}
$$

we get

$$
\vec{b}+\mu\left(-\vec{b}+\frac{1}{2} \vec{a}\right)=\lambda(\vec{a}+\vec{b})
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_262e89c92fe147e6116eg-4.jpg?height=279&width=642&top_left_y=620&top_left_x=842)

and by rearranging,

$$
\left(\frac{1}{2} \mu-\lambda\right) \vec{a}=(\lambda+\mu-1) \vec{b}
$$

The vectors $\vec{a}$ and $\vec{b}$ are not collinear, so from the previous equation, it follows that

$$
\begin{aligned}
& \frac{1}{2} \mu-\lambda=0 \\
& \lambda+\mu-1=0
\end{aligned}
$$

from which we get $\lambda=\frac{1}{3}, \mu=\frac{2}{3}$.

Therefore,

$$
\overline{A M}: \overline{M C}=\frac{1}{3}: \frac{2}{3}=1: 2, \quad \overline{D M}: \overline{M E}=2: 1
$$

## III year