Question

In $△ABC$, if $AD$ is the median, show that $A{B}^2+A{C}^2=2(A{D}^2+B{D}^2)$.

Answer 1

$AD$ is median.

Drop a perpendicular $AM$, then,

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

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

$A{B}^2+A{C}^2=B{M}^2+C{M}^2+2A{M}^2$

$A{B}^2+A{C}^2=(BD+DM{)}^2+(CD-MO{)}^2+2(A{D}^2-D{M}^2)$

$\Rightarrow A{B}^2+A{C}^2=B{D}^2+C{D}^2+D{M}^2+D{M}^2+2\left(BD\right)\left(DM\right)-2\left(DM\right)\left(CD\right)+2A{D}^2-2D{M}^2$

$\Rightarrow 2A{D}^2+(B{D}^2+C{D}^2)+2\left(DM\right)[BD-CD]$

AS, $BD=CD$

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