Question

prove that three times the sum of squares of sides of triangle is equal to the four time squares os sum of medians os triangle

Answer 1

Hey friend....

Here is your answer....

As we know,

the sum of the squares of two sides of a triangle is equal to twice the square of the median on the third side plus half the square of the third side.

Hence,

AB^2 + AC^ 2 = 2BD^ 2 + 2AD^ 2 = 2 × (½BC)^2 + 2AD^2

= ½ BC^2 + 2AD^2

∴ 2AB^2 + 2AC^ 2 = BC^2 + 4AD^2 → (1) Similarly,

we get

2AB^2 + 2BC^2 = AC^2 + 4BE^2 → (2)

2BC^2 + 2AC^2 = AB^2 + 4CF^2 → (3)

Adding (1) (2) and (3)

we get,

4AB^2 + 4BC^2 + 4AC^ 2 = AB>2 + BC^2 + AC^2 + 4AD^2 + 4BE^2 + 4CF^2

3(AB^2 + BC^2 + AC^2) = 4(AD^2 + BE^2 + CF^2)

Hence,

three times the sum of squares of the sides of a triangle is equal to four times the sum of squares of the medians of the triangle.

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