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In a ∆ ABC, AD ,BE and CF are three medians. Then the ratio (AD + BE + CF) :
(AB + AC + BC) is
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equal to 3 4 -
less than 3 4 -
greater than 3 4 -
equal to 1 2 - None of these
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Correct Option: E
As per the given in question , we draw a figure of triangle ABC 
In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it.
∴ AB² + AC² = 2 (AD² + BD²)
| ⇒ AB² + AC² = 2 |  | AD² + |  | |
| 4 |
⇒ 2(AB² + AC²) = 4AD² + BC²
Similarly,
2(AB² + BC²) = 4 BE² + AC²
2 (AC² + BC²) = 4 CF² + AB²
On adding all three, we get
4 (AB² + BC² + AC²) = 4 (AD² + BE² + CF²) + BC² + AC² + AB²
⇒ 3 (AB² + BC² + AC²) = 4 (AD² + BE² + CF²)
