SUBIECTUL III In triangle $A B C$ with sides $A B=c, A C=b, B C=a$, let $P$ be the intersection of the angle bisector $C E$ and the median $B D, E \in(A B), D \in(A C)$.

Determine the numbers $x$ and $y$ such that $\overrightarrow{P A}=x \cdot \overrightarrow{P B}+y \cdot \overrightarrow{P C}$.
## Solution.

$\Delta P A B \Rightarrow \overrightarrow{P A}=\overrightarrow{P B}+\overrightarrow{B A}$ and $\Delta P A C \Rightarrow \overrightarrow{P A}=\overrightarrow{P C}+\overrightarrow{C A}$

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Adding the relations, we get:

$$
2 \overrightarrow{P A}=\overrightarrow{P B}+\overrightarrow{P C}+\overrightarrow{B A}+\overrightarrow{C A}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_a02813c7718881642711g-3.jpg?height=363&width=542&top_left_y=618&top_left_x=1405)

We obtain $\frac{2 a+b}{2 a} \overrightarrow{B P}=\overrightarrow{B A}+\overrightarrow{A D}$.

Substituting into relation (1),

we get: $2 \overrightarrow{P A}=-\frac{b}{2 a} \overrightarrow{P B}+\overrightarrow{P C}+\frac{3}{2} \overrightarrow{C A}$, but $\overrightarrow{C A}=\overrightarrow{C P}+\overrightarrow{P A}$, so $\overrightarrow{P A}=-\frac{b}{a} \overrightarrow{P B}-\overrightarrow{P C}$, thus $x=-\frac{b}{a}, y=-1$