1. Prove this!

In a triangle ABC, BD and CE are the medians of a triangle, meet at centroid G.
Prove that BG=2GD.

2. Centroid of a triangle

Hello snigdha
Originally Posted by snigdha
In a triangle ABC, BD and CE are the medians of a triangle, meet at centroid G.
Prove that BG=2GD.
If you understand how you can use vectors in geometry, you'll find a vector proof just here.

But if you want a more 'traditional' proof, look at the diagram I've attached.

In the diagram, $E$ and $D$ are the mid-points of $AB$ and $AC$ respectively.

If we consider the area of $\triangle ABC$ when its base is $AB$, we see that:
area of $\triangle AEC = \tfrac12$ area of $\triangle ABC$
because the base ( $AE$) of $\triangle AEC$ is one-half of the base ( $AB$) of $\triangle ABC$, and the height of each triangle is the same.

Similarly, when we consider $AC$ as the base of $\triangle ABC$, we get:
$\triangle ABD = \tfrac12 \triangle ABC$
Therefore:
$\triangle AEC = \triangle ABD = \tfrac12 \triangle ABC$
With the colours I've used in the diagram, this is:
blue area + green area + yellow area = red area + yellow area + green area
So, if we subtract the common area - the quadrilateral $AEGD$ (yellow area + green area) - from each triangle, we get:
$\triangle BGE = \triangle CGD$
In colours:
red area = blue area
But we can also see that, because $\triangle BGE$ (red) and $\triangle AGE$ (yellow) have equal bases and the same height:
$\triangle BGE=\triangle AGE$
In colours:
red area = yellow area
Similarly:
$\triangle AGD=\triangle CGD$
In colours:
green area = blue area
Thus all four coloured triangles have the same area. Therefore
$\triangle AGB = \tfrac23\triangle ABD$
But these triangles have a common base $AB$. Therefore the height of $\triangle AGB = \tfrac23$ of the height of $\triangle ABD$. So, by similar triangles:
$BG = \tfrac23BD$

$\Rightarrow BG = 2 GD$
Tricky, isn't it? (It's much easier to use the vector method!)

3. Originally Posted by snigdha
In a triangle ABC, BD and CE are the medians of a triangle, meet at centroid G.
Prove that BG=2GD.
hi

first , prove that $\triangle AED$ is similar to $\triangle ABC$

Proof : $\angle A$ is a common angle for both triangles . We also see that AB=2AE and AC=2AD . Since the ratio of 2 corresponding sides are equal and the included angles are equal , proved then .

so $\frac{AE}{AB}=\frac{ED}{BC}$ ----1

Then , prove that $\triangle EDG$ is similar to $\triangle CBG$ .

Proof : Since $\angle AED =\angle ABC$ and $\angle ADE=\angle ACB$ , ED is parallel to BC .

hence , $\angle DEG=\angle BCE$ , $\angle EDG=\angle CBD$ (alternate angles) , and $\angle EGD=\angle BGC$ (opposite angles)

since , all corresponding angles are equal , hence proved .

so $\frac{DG}{BG}=\frac{ED}{BC}$ ---2

From 1 , since $\frac{AE}{AB}=\frac{1}{2}$ , so $\frac{ED}{BC}=\frac{1}{2}$ , and it follows that $\frac{DG}{BG}=\frac{1}{2}$
so BG=2GD