Question

Prove that 3 times the sum of the squares of the sides of a triangle is equal to 4 times the sum of the squares of its medians.. and is this question in the syllabus of 10th class students???
Because we really dont have Apollonius theorem????
Please help with the correct answer. I'll surely mark brainliest

Answer 1

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 AB^2 + AC^2 = 2BD^2 +
2AD^2
= 2 × (½BC)2 + 2AD^2
= ½ BC2 + 2AD^2

∴ 2AB^2 + 2AC^2 = BC^2 + 4AD^2→(1)

Similarly, we get

2AB^2 + 2BC^2 = AC^2 + 4BE2 → (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(AD2^ + 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.

Answer 2

Without Apollonius theorem ↓↓↓


From the diagram, AX ⊥BC and AD is the median. 

In ∆s AXB, & AXC, by Pythagoras Theorem, 

AB² = AX² + BX² -------- (i) 

 AC² = AX² + CX² -------- (ii) 

Adding (i) & (ii)

=> AB² + AC² = 2 AX² + BX² + CX² 

But BX = BD - DX and CX = DX + DC

==> DX + BD { as BD = DC } 

Therefore, BX² + CX² = (BD - DX)² + (DX + BD)² 

=> BX² + CX² = (BD² + DX² - 2 BD* DX) + (DX² + BD² + 2BD * DX) 

=> BX² + CX² = 2BD² + 2DX² 

Now,

AB² + AC² = 2 AX² + BX² + CX² => AB² + AC² = 2AX² + 2BD² + 2DX² 

=> AB² + AC² = 2(AX² + DX²) + 2BD² { since AX² + DX² = AD² } 

=> AB² + AC² = 2AD² + 2BD² ==> AB² + AC² = 2AD² + BC²/2 -------(iii) 

Similarly; BC² + AB² = 2BE² + AC²/2 -------- (iv) and 

AC² + BC² = 2CF² + AB²/2 ------------(v) 

Now adding (iii) , (iv) and (v) we get ----> 

2(AB² + BC² + AC²) = 2(AD² + BE² + CF²) + (1/2)(AB² + BC² + AC²) 

Now Multiplying throughout by 2, 

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

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

☺ Hope this Helps ☺