0
$\begingroup$

In $\Delta ABC$ Can any one give me a hint to Prove that the centroid $G$ divides $A$ and Mid point of $BC$ in the ratio $2$:$1$ Using only Plane Geometry.

$\endgroup$
1
$\begingroup$

Let $L$, $M$ and $N$ be the midpoints of $AB$, $BC$ and $AC$ and $G$ be the centroid.

  1. The median divides the area of a triangle into 2 parts of equal area.
  2. $S_{\triangle ABM} = S_{\triangle ACM} $ since $AM$ is median of $S_{\triangle ABC}$.
  3. $S_{\triangle GBM} = S_{\triangle GCM} $ since $GM$ is median of $S_{\triangle GBC}$.
  4. Thus by subtraction, $S_{\triangle ABG} = S_{\triangle ACG} $.
  5. By symmetry we have $S_{\triangle ABG} = S_{\triangle ACG} = S_{\triangle BCG} $.
  6. So the area of $\triangle BCG$ is one third of the area of the whole triangle $\triangle ABC$. $S_{\triangle ABC} = 3\cdot S_{\triangle BCG}$
  7. So $AM = 3 \cdot GM$. And we can conclude now.
$\endgroup$
1
$\begingroup$

Let $A_1,B_1$ be midpoints of $BC, AC$, respectively. Then $AA_1\cap BB_1= G$.

$\triangle CB_1A_1\sim \triangle CBA$, because $\angle C$ is common and $\frac{CB_1}{CA}=\frac{CA_1}{CB}=\frac{1}{2}$,

so $\frac{A_1B_1}{AB}=\frac{1}{2}$ and $A_1B_1||AB$ and $\angle BAA_1=\angle AA_1B_1$ and $\angle ABB_1=\angle A_1B_1B$,

so $\triangle ABG\sim\triangle A_1B_1G$ with ratio $\frac{A_1B_1}{AB}=\frac{1}{2}$, so $\frac{BG}{GB_1}=\frac{AG}{GA_1}=\frac{1}{2}$.

$\endgroup$
1
$\begingroup$

Alternative proof, a bit more interesting (see picture):

$\frac{CB}{CF}=\frac{CB_1}{CA}$, so $BB_1 || FA$, so $\frac{A_1B}{A_1F}=\frac{A_1G}{A_1A}=\frac{1}{3}$.

enter image description here

$\endgroup$
1
$\begingroup$

Let $AA_1$ and $BB_1$ be two medians, and $G$ their intersection. Denote by $A_2$ respectively $B_2$ the midpoints of $AG$ respectively $BG$.

Then, since $A_1B_1$ is midline in $CBA$ and $A_2B_2$ is midline in $GBA$ we have that both are parallel to $AB$ and half of $AB$. This shows that the oposite sides $A_1B_1$ an $A_2B_2$ in the quadrilateral $A_1B_1A_2B_2$ are parallel and equal. This shows that $A_1B_1A_2B_2$ is parallelogram.

The claim follows from the fact that the diagonals of this parallelogram half eachother.

$\endgroup$
  • $\begingroup$ Can the downvoter explain what is wrong with this solution? $\endgroup$ – N. S. Jun 11 '15 at 3:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.