Question

If G is the centroid of the triangle ABC and if G' is
the centroid of another triangle A'B'C' then prove
that: AA' + BB' + CC' = 3GG'​

Answer 1

Answer:

$\textbf{Concept used:}$

$\boxed{\begin{minipage}{7cm} The\:position\:vector\:of\:Centroid\:of\:\triangle\:ABC\:is\\ \\\,\overrightarrow{OG}=\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3} \end{minipage}}$

$\text{Let O be the origin}$

$\text{Given:}$

$\text{G and G' be centroids of }\trianlge\,ABC\text{ and }\trianlge\,A'B'C'$

$\text{Then,}$

$\bf\overrightarrow{OG}=\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}\,\text{ and }\overrightarrow{OG'}=\frac{\overrightarrow{OA'}+\overrightarrow{OB'}+\overrightarrow{OC'}}{3}$

$\text{Now}$

$\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}$

$=(\overrightarrow{OA'}-\overrightarrow{OA})+(\overrightarrow{OB'}-\overrightarrow{OB})+(\overrightarrow{OC'}-\overrightarrow{OC})$

$=(\overrightarrow{OA'}+\overrightarrow{OB'}+\overrightarrow{OC'})-(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})$

$=3\,\overrightarrow{OG'}-3\,\overrightarrow{OG}$

$=3(\overrightarrow{OG'}-\,\overrightarrow{OG})$

$=3\,\overrightarrow{GG'}$

$\implies\boxed{\bf\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=3\,\overrightarrow{GG'}}$

Answer 2

Answer:

According to the question, G is the center of the 1st triangle ABC and G' is

the center of 2nd triangle A'B'C'

Let, O is the origin  of the triangle

then, For the first triangle -

Equation (i)  →  OG  = ( OA + OB + OC ) / 3

and similarly,  

Equation (ii)  →  OG` = ( OA` + OB` + OC` ) / 3  

   

subtract  equation (ii) from equation (i)

OG -OG' =  ( OA +OB +OC ) / 3 - ( OA` + OB` + OC` ) / 3    

⇒ GG`  = ( AA` + BB` + CC` ) / 3

⇒  AA`+BB`+CC`= 3GG` ( proved )