Question

If AD, BE and CF are the medians of a $△ABCthen(A{D}^2+B{E}^2+C{F}^2):(B{C}^2+C{A}^2+A{B}^2)$ is equal to

• A. 4:3
• B. 3:2
• C. 3:4
• D. 2:3

Answer 1

The correct option is C

3:4

$A{B}^2+A{C}^2=2(A{D}^2+B{D}^2)$
$\Rightarrow \frac{c^2+b^2}{2}-\frac{a^2}{4}=A{D}^2$
Similarly, $\frac{a^2+b^2}{2}-\frac{c^2}{4}=C{F}^2$ and $\frac{a^2+c^2}{2}-\frac{a^2+c^2}{4}=B{E}^2$
Adding $a^2+b^2+c^2-\left(\frac{a^2+b^2+c^2}{4}\right)=A{D}^2+B{E}^2+C{F}^2$
$\Rightarrow (A{D}^2+B{E}^2+C{F}^2):(a^2+b^2+c^2)=3:4$