Question

If in a triangle abc, the medians AD,BE andCF meet atG, then which of the following is true​

Answer 1

Answer:

Firstly , We draw a figure of triangle ABC whose the medians AD, BE and CF meet at G ,

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 + BC²/4)

⇒ 2(AB² + AC²) = 4 AD² + 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)

Again,

AB + AC > 2AD

AB + BC > 2BE

BC + AC > 2CF

On adding , we get

∴ 2(AB + BC + AC) > 2(AD + BE + CF)

⇒ AB + BC + AC > AD + BE + CF

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