Question

BL and CM are medians of a triangle ABC right angled at A. Prove that
$4({BL}^2+{CM}^2)=5{BC}^2$. [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark
Proof : 1 Mark

BL and CM are medians of the $\Delta ABC$ in which $\angle A={90}^{\circ }$.





In $\Delta ABC,$

${BC}^2={AB}^2+{AC}^2.......\left(1\right)$ [Pythagoras Theorem]

In $\Delta ABL,$

${BL}^2={AL}^2+{AB}^2$ [Pythagoras Theorem]

$\Rightarrow {BL}^2={\left(\frac{AC}{2}\right)}^2+{AB}^2$ [L is the mid-point of AC]

$\Rightarrow {BL}^2=\frac{{AC}^2}{4}+{AB}^2$

$\Rightarrow 4{BL}^2={AC}^2+4{AB}^2.......\left(2\right)$

From $\Delta CMA,$

${CM}^2={AC}^2+{AM}^2$

or, ${CM}^2={AC}^2+{AM}^2$

[M is the mid-point of AB]

${CM}^2={AC}^2+\frac{{AB}^2}{4}$

$4{CM}^2=4{AC}^2+{AB}^2......\left(3\right)$

Adding (2) and (3), we have

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

$\Rightarrow 4({BL}^2+{CM}^2)=5{BC}^2$ [From (1)]