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

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Solution

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

BL and CM are medians of the $Δ A B C$ in which $∠ A = 90^{\circ}$ .

In $Δ A B C,$

${B C}^{2} = {A B}^{2} + {A C}^{2} . . . . . . . (1)$ [Pythagoras Theorem]

In $Δ A B L,$

${B L}^{2} = {A L}^{2} + {A B}^{2}$ [Pythagoras Theorem]

$\Rightarrow {B L}^{2} = {(\frac{A C}{2})}^{2} + {A B}^{2}$ [L is the mid-point of AC]

$\Rightarrow {B L}^{2} = \frac{{A C}^{2}}{4} + {A B}^{2}$

$\Rightarrow 4 {B L}^{2} = {A C}^{2} + 4 {A B}^{2} . . . . . . . (2)$

From $Δ C M A,$

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

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

[M is the mid-point of AB]

${C M}^{2} = {A C}^{2} + \frac{{A B}^{2}}{4}$

$4 {C M}^{2} = 4 {A C}^{2} + {A B}^{2} . . . . . . (3)$

Adding (2) and (3), we have

$4 ({B L}^{2} + {C M}^{2}) = 5 ({A C}^{2} + {A B}^{2})$

$\Rightarrow 4 ({B L}^{2} + {C M}^{2}) = 5 {B C}^{2}$ [From (1)]

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