Question

AD is the median of $\Delta ABC$ and $\angle BAD={90}^{\circ }$

Express BC in terms of AB and AC.

Answer 1

length of median AD formula:

$AD=\frac{1}{2}\sqrt{2({AB}^2+{AC}^2)-{BC}^2}-\left(1\right)$

using pythagoras theorem in$△ABC$

${AB}^2+{AD}^2={BD}^2-\left(2\right)$

using(1) in (2)

${AB}^2+\frac{1}{4}\left[2\right({AB}^2+{AC}^2)-{BC}^2]={BD}^2$

$⟹4{AB}^2+2{AB}^2+2{AC}^2-{BC}^2={BD}^2$

$⟹6{AB}^2+2{AC}^2-{BC}^2={BD}^2$

$\because BD=\frac{BC}{2}$ ($\because$ AD is a median)

$⟹6{AB}^2+2{AC}^2-{BC}^2=\frac{{BC}^2}{4}$

$⟹24{AB}^2+8{AC}^2-4{BC}^2={BC}^2$

$⟹{BC}^2=\frac{24{AB}^2+8{AC}^2}{5}$

$⟹{BC}^2=\frac{8}{5}[3{AB}^2+{AC}^2]$