Question

Let AD be a median of the $△ABC.$ If AE and AF are medians of the triangle ABD and ADC, respectively, and $BD=\frac{a}{2},AD=m_1,AE=m_2,AF=m_3,$ then $\frac{a^2}{8}$ is equal to:

• A. $m_2^2+m_3^2-2m_1^2$
• B. $m_1^2+m_2^2-2m_3^2$
• C. $m_1^2+m_3^2-2m_2^2$
• D. none of these

Answer 1

The correct option is A $m_2^2+m_3^2-2m_1^2$

In $\Delta ABC$

By Apollonius's theorem

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

$\Rightarrow A{B}^2+A{C}^2=2(A{D}^2+{\left(\frac{a}{2}\right)}^2)$
$\Rightarrow A{B}^2+A{C}^2-\frac{a^2}{2}=2m_1^2$ ....(1)
Similarly in $\Delta ABD$: $A{B}^2+A{D}^2-\frac{{BC}^2}{8}=2A{E}^2$
$\Rightarrow A{B}^2+m_1^2-\frac{a^2}{8}=2m_2^2$ ....(2)
and in $\Delta ACD$: $A{D}^2+A{C}^2-\frac{{BC}^2}{8}=2A{F}^2$
$\Rightarrow A{C}^2+m_1^2-\frac{a^2}{8}=2m_3^2$ ....(3)

now adding (2) & (3) followed by subtracting (1), we get

$\frac{a^2}{4}+2{m_1}^2=2m_2^2+2m_3^2-2m_1^2$
Therefore, $\Rightarrow \frac{a^2}{8}=m_2^2+m_3^2-2m_1^2$