Question

. Prove that three times the sum of the squares of
the sides of a triangle is equal to four times the
sum of squares of the medians of that
triangle.
(HOTS)​

Answer 1

Answer:Hope it helps

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Step-by-step explanation:

Apollonius theorem states that 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 AB2 + AC 2 = 2BD 2 + 2AD 2

= 2 × (½BC)2 + 2AD2

= ½ BC2 + 2AD2

∴ 2AB2 + 2AC 2 = BC2 + 4AD2 → (1)

Similarly, we get

2AB2 + 2BC2 = AC2 + 4BE2 → (2)

2BC2 + 2AC2 = AB2 + 4CF2 → (3)

Adding (1) (2) and (3), we get

4AB2 + 4BC2 + 4AC 2 = AB2 + BC2 + AC2 + 4AD2 + 4BE2 + 4CF2

3(AB2 + BC2 + AC2) = 4(AD2 + BE2 + CF2)

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.