Question

If BL and CM are the medians of $\Delta ABC$ right angled at A. The value of $4(B{C}^2+C{M}^2)-5B{C}^2$ is:

Answer 1

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

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

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

$=\frac{1}{4}[{AB}^2+{AC}^2+{4BC}^2]$

$4({BL}^2+{CM}^2)={AB}^2+{AC}^2+{4BC}^2$

$Now,{AB}^2+{AC}^2={BC}^2(\because triangleisrightangledatA)$

$\therefore pythagorastheorem$

$4({BL}^2+{CM}^2)=5{BC}^2$

$hence,4({BL}^2+{CM}^2)=5{BC}^2$

$proved$