Question

Three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle.

• A. True
• B. False

Answer 1

The correct option is A True

Since in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it.

Therefore, taking AD as the median bisecting side BC, we have

$A{B}^2+A{C}^2=2(A{D}^2+B{D}^2)$

$A{B}^2+A{C}^2=2[A{D}^2+(\frac{BC}{2}{)}^2]$

$A{B}^2+A{C}^2=2(A{D}^2+\frac{B{C}^2}{4})$

$2(A{B}^2+A{C}^2)=(4A{D}^2+B{C}^2)$ .....(1)

Similarly, by taking BE and CF, respectively, as the medians,

$2(A{B}^2+B{C}^2)=(4B{E}^2+A{C}^2)$ .......(2)

and $2(A{C}^2+B{C}^2)=(4C{F}^2+A{B}^2)$ .....(3)

Adding 1, 2 and 3, we get,

$4(A{B}^2+B{C}^2+A{C}^2)=4(A{D}^2+B{E}^2+C{F}^2)+(B{C}^2+A{C}^2+A{B}^2)$

Therefore, $3(A{B}^2+B{C}^2+A{C}^2)=4(A{D}^2+B{E}^2+C{F}^2)$