5.24. In parallelogram $A B C D$ it is given: $M \in B C$ and $B M: M C=1: 2$; $N \in D C, D N: N C=1: 2 ; \overline{A M}=\bar{a} ; \overline{A N}=\bar{b}$. Express the vectors $\overline{A B}, \overline{A D}, \overline{M N}$ and $\overline{B D}$ in terms of $\bar{a}$ and $\bar{b}$.
5.24. Given that $\overline{A M}=\bar{a}, \overline{A N}=\bar{b}$ (Fig. 5.14) and, therefore, $\overline{M N}=\overline{A N}-\overline{A M}=\bar{b}-\bar{a}. \quad$ Since

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-609.jpg?height=290&width=496&top_left_y=1151&top_left_x=44)

Fig. 5.14 $\frac{B M}{M C}=\frac{1}{2}, \frac{D N}{N C}=\frac{1}{2}$, so $\quad \frac{M C}{B C}=\frac{2}{3}$, $\frac{N C}{C D}=\frac{2}{3}$ and angle $C$ is common to triangles $N C M$ and $D C B$. Therefore, $\triangle N C M \sim \triangle D C B$, from which

$$
\frac{M N}{B D}=\frac{2}{3} \quad \text { and } \quad \overline{B D}=\frac{3}{2} \overline{M N}, \quad \text { i.e. }
$$

$\overline{B D}=\frac{3}{2}(\bar{b}-\bar{a})$. Let $\overline{A D}=x, \overline{A B}=\bar{y}$. We have $\overline{A D}-\overline{A B}=\overline{B D}$, or

$$
\bar{x}-\bar{y}=\frac{3}{2}(\bar{b}-\bar{a})
$$

Further,

$$
\overline{A B}+\overline{B M}=\overline{A M}, \text { or } \bar{y}+\frac{1}{3} \bar{x}=\bar{a}
$$

It remains to solve the system of equations (1) and (2); as a result, we get $\overline{A D}=\frac{9}{8} \bar{b}-\frac{3}{8} \bar{a}, \overline{A B}=\frac{9}{8} \bar{a}-\frac{3}{8} \bar{b}$.

## Answer:

$$
\overline{A B}=\frac{9}{8} \bar{a}-\frac{3}{8} \bar{b}, \overline{A D}=\frac{9}{8} \bar{b}-\frac{3}{8} \bar{a}, \overline{M N}=\bar{b}-\bar{a}, \overline{B D}=\frac{3}{2} \bar{b}-\frac{3}{2} \bar{a}
$$