Question

If $AD,BE$ and $CF$ are the medians of a $\Delta ABC,$ then evaluate $(A{D}^2+B{E}^2+C{F}^2):(B{C}^2+C{A}^2+A{B}^2)=$

• A. $3:4$
• B. $4:3$
• C. $5:3$
• D. $4:1$

Answer 1

The correct option is A $3:4$

Given, $AD,BE$ and $CF$ are the medians of a $\Delta ABC$.
$\Rightarrow A{B}^2+A{C}^2=2(A{D}^2+B{D}^2)$
$\Rightarrow A{B}^2+A{C}^2=2A{D}^2+\frac{B{C}^2}{2}$
$\Rightarrow 2A{D}^2=A{B}^2+A{C}^2-\frac{B{C}^2}{2}$ -----(1)
Similarly,
$2B{E}^2=B{C}^2+B{A}^2-\frac{A{C}^2}{2}$ -----(2)
$2C{F}^2=C{A}^2+C{B}^2-\frac{A{B}^2}{2}$ -----(3)
Adding equation 1,2 and 3, we get
$2(A{D}^2+B{E}^2+C{F}^2)=\frac{3}{2}(A{B}^2+B{C}^2+C{A}^2)$
$\therefore (A{D}^2+B{E}^2+C{F}^2):(A{B}^2+B{C}^2+C{A}^2)=3:4$