Question

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

• A. $4:3$
• B. $3:2$
• C. $3:4$
• D. $2:3$

Answer 1

Answer: (C)

$3:4$

In $△ABC,AD,BE$ and $CF$ are medians to respective sides.

So,

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

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

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

By Pythagoras theorem in $△AFC$

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

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

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

$4A{C}^2=4C{F}^2+A{B}^2$ ....... $\left(i\right)$

By Pythagoras theorem in $△BCE$

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

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

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

$4B{C}^2=4B{E}^2+A{C}^2$ ........ $\left(ii\right)$

By Pythagoras theorem in $△ADB$

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

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

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

$4A{B}^2=4A{D}^2+B{C}^2$ ....... $\left(iii\right)$

Adding $\left(i\right),\left(ii\right)$ and $\left(iii\right),$ we get

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

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

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

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

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