Answer
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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.”
Complete step-by-step solution:
Let us first of all draw the triangle named as ABC with its medians and centroid.
![seo images](https://www.vedantu.com/question-sets/6ba27e68-33f0-4c9f-bda0-048b8693cfe2458374247701120354.png)
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:-
![seo images](https://www.vedantu.com/question-sets/3be0f61d-7e82-445e-b3ca-e3f9baab058c5346931408598408759.png)
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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