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chengbin · #1 July 2nd, 2009, 08:09 PM

Given $\triangle$ 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 $\overrightarrow {PD}+\overrightarrow {PE}+\overrightarrow {PF}=\overrightarrow {PA}+\overrightarrow {PB}+\overrightarrow {PC}$ .

(solution in the book)

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

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

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

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

Why? How do they get that?