Question

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

• A. $\frac{3}{4}$
• B. $\frac{3}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{3}{4}$


Let the sides of $\Delta ABC$ are $a,b,c$
$\therefore BC=a,CA=b$ and $AB=c$
Again $AD,BE$ and $CF$ are the medians of $\Delta ABC$
$\therefore AD=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$
$BE=\frac{1}{2}\sqrt{2a^2+2c^2-b^2}$
$CF=\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$
$\therefore A{D}^2+B{E}^2+C{F}^2=\frac{1}{4}[2b^2+2c^2-a^2+2a^2+2c^2-b^2+2a^2+2b^2-c^2]$
$=\frac{1}{4}[3a^2+3b^2+3c^2]=\frac{3}{4}[a^2+b^2+c^2]$
$\Rightarrow A{D}^2+B{E}^2+C{F}^2=\frac{3}{4}[B{C}^2+C{A}^2+A{B}^2]$
$\therefore \frac{A{D}^2+B{E}^2+C{F}^2}{B{C}^2+C{A}^2+A{B}^2}=\frac{3}{4}$