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

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Solution

$B L is median$
$\Rightarrow$ $A L = C L = \frac{1}{2} A C \to (1)$

$CM is median$
$\Rightarrow$ $AM = MB = \frac{1}{2} A B \to (2)$

$I n Δ B A C$
${(B C)}^{2} = {(A B)}^{2} + {(A C)}^{2}$

$I n Δ B A C$
${(B L)}^{2} = {(A B)}^{2} + {(\frac{A C}{2})}^{2}$

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

$I n Δ MAC,$
${(C M)}^{2} = {(A M)}^{2} + {(A C)}^{2}$

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

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

$N O W, {(B C)}^{2} = {(A B)}^{2} + {(A C)}^{2} \to (1)$

$4B C^{2} = 4 {(A B)}^{2} + {(A C)}^{2} \to (2)$

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

$A D D (2) & (3)$

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

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

Hence proof

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Similar questions

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

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]

B L

C M

Δ A B C

A

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

Δ

∠

= 90^{o}

Δ

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

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