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A NEW G E O M E T R I C OBJECT

We will work within the framework of affine geometry, using vectors, on the following configuration :

Object data : a triangle $ABC$ ;

a cevian $AD$ ;

a point $P\;\; on\;\; AD$ ;

fixing point $D \;\;on\;\; BC\;\;:\;\frac{\overline{DB}}{\overline{DC}}=-m$

fixing point $P\;\;\;:\;\;\frac{\overline{PA}}{\overline{PD}}=-n$

Conclusion on the Object : $(1+m)\cdot \overrightarrow{PA}+n\cdot \overrightarrow{PB}+m\cdot n\cdot \overrightarrow{PC}=\overrightarrow{0}\;\;\;(1)$

.

Proof CiP

We will use the formulas established previously—here or here.

With points $P, \;B,\; C\; and\; D\;:\;\frac{\overline{DB}}{\overline{DC}}=-m\;\Rightarrow$

$\Rightarrow\;\;\;\overrightarrow{PD}=\frac{1}{1+m}\cdot \overrightarrow{PB}+\frac{m}{1+m}\cdot \overrightarrow{PC} \tag{2}$

But $\frac{\overline{PA}}{\overline{PD}}=-n\;\;\Rightarrow\;\overrightarrow{PA}=-n\cdot \overrightarrow{PD}\;\Rightarrow\;\overrightarrow{PD}=-\frac{1}{n}\cdot \overrightarrow{PA}$ which, when substituted

into (2) , gives us

$-\frac{1}{n}\cdot \overrightarrow{PA}=\frac{1}{1+m}\cdot \overrightarrow{PB}+\frac{m}{1+m}\cdot \overrightarrow{PC}$                                                                            

and we immediately obtain (1).

$\blacksquare\;\;\blacksquare$

If, in particular, $if\;P=G$ is the centroid of triangle $ABC$, then $m=1\;,\;n=2$

and we have the well-known formula :

$\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}$