Given $\displaystyle \triangle$ABC, where points D, E, F are the midpoints of the sides BC, CA, and AB, respectively, and given an arbitrary point P, show that $\displaystyle \overrightarrow {PD}+\overrightarrow {PE}+\overrightarrow {PF}=\overrightarrow {PA}+\overrightarrow {PB}+\overrightarrow {PC}$.

(solution in the book)

Letting $\displaystyle \overrightarrow {a}, \overrightarrow {b}$, and $\displaystyle \overrightarrow {c}$ be the position vectors of the vertices A, B, and C, respectively (originating from point P)

$\displaystyle \overrightarrow {PD} = \frac {1}{2}(\overrightarrow {b}+\overrightarrow {c})$

$\displaystyle \overrightarrow {PE} = \frac {1}{2}(\overrightarrow {a}+\overrightarrow {c})$

$\displaystyle \overrightarrow {PF} = \frac {1}{2}(\overrightarrow {a}+\overrightarrow {b})$

Why? How do they get that?