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.
​

Answer 1

AD is the median

In triangle AXB & AXC, by pythagoras theorem,

ABP = AX2 + BX2

and AC2 = AX+ CX2

Adding (1) & (2)

AB2 + AC2 = 2AX2 + BX? + CX?

but BX = BD - DX and CX= DX +

DC = BD + DX

. BX? + CX? = (BD DX)? +

(DX + BD)

BX2 + CX2 (BD2 + DX? - 2BD x

DX) + (DX² + BD2 – 2BD x DX)

BX? + CX? = 2BD? + 2 DX?

Now, AB? + AC2 CX? 2AX? + BX? +

AB? + AC2 = 2AX2 + 2BD2 + 2DX2

AB? + AC? = 2(AX? + DX?) + 2BD? :: AX? + DX? = AD? =

AB? + AC? = 2AD? + 2BD² =

AB? + AC? = 2AD? = BC2 2 + (3)

similarly, BC2 + ABP = 2BE2 +

(4)

AC? + BC2 2CF2 + AB2 2

Adding (3),(4) and (5)

2(AB? + BC? + AC?) = 2(AD? +

BE2 1 +CF2)+(ABP + BC2 + AC?) 2

multiplying throughout by 2,

4(AB? + BC? + AC?) = 4(AD? + = BE? + CF2) + (AB? + BC? + AC?)

3(AB? + BC? + AC?) = 4(AD? + BE+ CF2)

Hence Proved.