Question

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

• A. $3:4$
• B. $2:3$
• C. Minimum value of $\frac{3}{8}({\tan }^2\theta +{\cot }^2\theta )$
• D. Minimum value of $\frac{1}{3}({\tan }^2\theta +{\cot }^2\theta )$

Answer 1

The correct option is A $3:4$

If in $△ABC,AD,BE$ and $CF$ are median and altitude to respective sides, then,

$BD=DC=\frac{1}{2}BC$

$AE=EC=\frac{1}{2}AC$

And, $AF=FB=\frac{1}{2}AB$

Now in $△AFC$, by pythagoras theorem,

$A{C}^2=C{F}^2+A{F}^2$

$A{C}^2=C{F}^2+(\frac{1}{2}AB{)}^2$

$A{C}^2=C{F}^2+\frac{1}{4}A{B}^2$

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

In $△BCE$, by pyhtagoras theorem,

$B{C}^2=B{E}^2+E{C}^2$

$B{C}^2=B{E}^2+(\frac{1}{2}AC{)}^2$

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

In $△ADB$, by pythagoras theorem,

$A{B}^2=A{D}^2+B{D}^2$

$A{B}^2=A{D}^2+(\frac{1}{2}BC{)}^2$

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

Adding equation 1,2, and 3, we get,

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

$\frac{A{D}^2+B{E}^2+C{F}^2}{A{B}^2+B{C}^2+A{C}^2}=\frac{3}{4}$