Let A,B,C be the points of triangle and P inside of triangle. How can I prove using vectors that: ( A P → + C P → + B P → ) = 0 → only if P is centroid

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Let A,B,C be the points of triangle and P inside of triangle. How can I prove using vectors that:  (                    A        P            →        +                    C        P            →        +                    B        P            →        )  =            0      →      only if P is centroid

Kelsie Marks · Accepted Answer

I assume that you are using the definition of the centroid of a triangle as the intersection of its medians. The essential idea is to use the following important property of vectors: for any points X, Y and Z,                    X        Z            →        =                    X        Y            →        +                    Y        Z            →        .(This is easy to prove by the definitions.) Now, note that                                                        A              P                        →                          +                                            C              P                        →                          +                                            B              P                        →                          =        0                                    ⟺                                                    A              P                        →                          +                                            A              P                        →                          −                                            A              C                        →                          +                                            A              P                        →                          −                                            A              B                        →                          =        0                                                  ⟺                                                    A              P                        →                          =                                                                              A                  B                                →                                      +                                                            A                  C                                →                                              3                                                          ⟺                                                    A              P                        →                          =                              2                                                            A                  D                                →                                              3                ,            where   D is the midpoint of side   B  C. This proves that   P lies on the median   A  D; by symmetry, we know that   P also lies on the medians   B  E and   C  F (where E, F are the midpoints of CA, AB respectively), so P is at the intersection of medians of the triangle.In other words, your statement implies that P is the centroid. I&#039;ll leave it as an exercise to you to prove the other direction: that is, if P is the intersection of medians of the triangle, the equation is satisfied.

garkochenvz · Accepted Answer

Let             a      →        ,            b      →        ,            c      →        be position vectors of points A,B,C. By definition, the position vector of the centroid            p      →        =                              a          →                    +                        b          →                    +                        c          →                      3  But then                    A        P            →        +                    B        P            →        +                    C        P            →        =  (            p      →        −            a      →        )  +  (            p      →        −            b      →        )  +  (            p      →        −            c      →        )    =  3            p      →        −            a      →        −            b      →        −            c      →          =  0

Let A,B,C be the points of triangle and P inside of triangle. How can I prove using vectors that: (vec(AP)+vec(CP)+vec(BP))=0⃗

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