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**Hint:**We will first draw a triangle and its median to get the centroid. After that, we will extend the end such that we get an inverted triangle downwards and then by the use of mid – point theorem get the required answer.

**Complete step-by-step solution:**

Let us first of all draw the triangle named as ABC with its medians and centroid.

Now we will extend AD to K such that AG = GK as well as join BK and CK.

We will then get a figure looking as follows:-

Now look at \[\vartriangle ABK\]:

Since, F and G are mid points of AB and AK respectively (Because we extended AB such that AG = GK).

$\therefore $ FG || BK (By the mid – point theorem)

Now since FG is a part of FC, therefore we will get:-

$\therefore $ GC || BK …………………(1)

Now look at \[\vartriangle ACK\]:

Since, E and G are mid points of AC and AK respectively (Because we extended AB such that AG = GK).

$\therefore $ GE || CK (By the mid – point theorem)

Now since GE is a part of BE, therefore we will get:-

$\therefore $ BG || CK …………………(2)

Now using (1) and (2), we get that:-

BGCK is a parallelogram.

We know that diagonals in a parallelogram bisect each other.

$\therefore $ GD = DK ……………….(3)

And we already have AG = GK

We can write it as follows:-

$ \Rightarrow $ AG = GD + DK

Now, on using (3), we will get the following expression:-

$ \Rightarrow $ AG = 2GD

Now, we can write this as following expression:-

$ \Rightarrow \dfrac{{AG}}{{GD}} = \dfrac{2}{1}$

$\therefore $ the centroid of the triangle divides each of its median in the ratio 2:1.

**Hence, the correct option is (B).**

**Note:**Now, let us understand the mid – point theorem which we used in the solution of the question above.

The mid – point theorem states that “the line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

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