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Question

Prove that three times the sum of the square of the sides of a triangle is equal to four times the sum of the square of the medians of the triangle.

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Solution

From the diagram
BC and AD is the median.
In AXBandAXC By Pythagoras Theorem,
AB2=AX2+BX2(i)
And
AC2=AX2+CX2(ii)
Adding (i) and (ii) we get
AB2+AC2=2AX2+BX2+CX2
But, BX=BDDX and CX=DX+DC
DX+BD since BD=DC
BX2+CX2=(BDDX)2+(DX+BD)2
BX2+CX2=(BD2+DX22BD×DX)+(DX2+BD2+2BD×DX)

BX2+CX2=2BD2+2DX2
Now
AB2+AC2=2AX2+BX2+CX2
AB2+AC2=2AX2+2BD2+2DX2
AB2+AC2=2(AX2+DX2)+2BD2{AX2+DX2=AD2}
AB2+AC2=2AD2+2BD2
Now,
AB2+AC2=2AD2+BC22(iii)
Similarly
BC2+AB2=2BE2+AC22(iv)
AC2+BC2=2CF2+AB22(v)

Now adding (iii),(iv)and(v) we get
2(AB2+BC2+AC2)=2(AD2+BC2+CE2)+12(AB2+BC2+AC2)
On multiplying through 2 we get
4(AB2+BC2+AC2)=4(AD2+BC2+CE2)+(AB2+BC2+AC2)
3(AB2+BC2+AC2)=4(AD2+BC2+CE2)
Proved.

1202500_1282639_ans_e8c0b24b62594459938afd010cc065d9.JPG

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