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, $\displaystyle E$ and $\displaystyle D$ are the mid-points of $\displaystyle AB$ and $\displaystyle AC$ respectively.
If we consider the area of $\displaystyle \triangle ABC$ when its base is $\displaystyle AB$, we see that: area of $\displaystyle \triangle AEC = \tfrac12$ area of $\displaystyle \triangle ABC$
because the base ($\displaystyle AE$) of $\displaystyle \triangle AEC$ is one-half of the base ($\displaystyle AB$) of $\displaystyle \triangle ABC$, and the height of each triangle is the same.
Similarly, when we consider $\displaystyle AC$ as the base of $\displaystyle \triangle ABC$, we get: $\displaystyle \triangle ABD = \tfrac12 \triangle ABC$
Therefore:$\displaystyle \triangle AEC = \triangle ABD = \tfrac12 \triangle ABC$
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 $\displaystyle AEGD$ (yellow area + green area) - from each triangle, we get:$\displaystyle \triangle BGE = \triangle CGD$
In colours:red area = blue area
But we can also see that, because $\displaystyle \triangle BGE$ (red) and $\displaystyle \triangle AGE$ (yellow) have equal bases and the same height:$\displaystyle \triangle BGE=\triangle AGE$
In colours:red area = yellow area
Similarly:$\displaystyle \triangle AGD=\triangle CGD$
In colours:green area = blue area
Thus all four coloured triangles have the same area. Therefore$\displaystyle \triangle AGB = \tfrac23\triangle ABD$
But these triangles have a common base $\displaystyle AB$. Therefore the height of $\displaystyle \triangle AGB = \tfrac23$ of the height of $\displaystyle \triangle ABD$. So, by similar triangles:$\displaystyle BG = \tfrac23BD$
$\displaystyle \Rightarrow BG = 2 GD$
Tricky, isn't it? (It's much easier to use the vector method!)
Grandad