Question

prove that three times the sum of the squares of the side of
a traingle is equal to four times the sum of the squares of the squares of the medians of the traingle

Answer 1

To Prove that: 3(AB^2+BC^2+CA^2) = 4(AD^2+BE^2+CF^2CF^2)

Construction: Draw AM perpendicular to BC,

Taking AD=h, BM=x-a, MD=a, BD=x, DC=x

Proof: AB^2 + AC^2 = {h^2 + (x-a)^2} + (h^2 + (x+a)^2} (AB^2= AM^2 + BM^2 and AC^2= AM^2 + MC^2)

= 2h^2 + 2x^2 + 2a^2

= 2(h^2 + a^2) + (2x)^2/2

= 2AD^2 + 1/2BC^2 .......... (1)

Similarly, AB^2 + BC^2 = 2BE^2 + 1/2AC^2 .......... (2)

and, BC^2 + AC^2 = 2CF^2 + 1/2AB^2 ..........(3)

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

2(AB^2 + BC^2+ CA^2) = 2(AD2 + BE2 + CF2) + 1/2(AB^2 + BC^2+ CA^2)

2(AB^2 + BC^2+ CA^2) = {4( AD2 + BE2 + CF2) + (AB^2 + BC^2+ CA^2)} /2

4(AB^2 + BC^2+ CA^2) - (AB^2 + BC^2+ CA^2 ) = 4( AD2 + BE2 + CF2)

3(AB^2 + BC^2+ CA^2) = 4( AD2 + BE2 + CF2) Hence, Proved.