BL and CM are medians of a triangle ABC right angled at A. Prove that $B L^{2} + C M^{2} = k B C^{2}$ and find the value of k.

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Solution

The triangle is right angled at A; BL and CM are medians.

Apply Pythagoras theorem to $Δ A B C$

$B C^{2} = A B^{2} + A C^{2}$
In triangle ABL,
$B L^{2} = A L^{2} + A B^{2}$
$B L^{2} = {(\frac{A C}{2})}^{2} + A B^{2}$ $(L is the midpoint of AC)$
$B L^{2} = (\frac{A C^{2}}{4}) + A B^{2}$
$4 B L^{2} = A C^{2} + 4 A B^{2} - - - - - - - - - - - - - - - - - - - - - - - - - - 1$
Similarly in triangle CMA we get
$4 C M^{2} = 4 A C^{2} + A B^{2} - - - - - - - - - - - - - - - - - - - - - - - - 2$
Adding 1 and 2, we get:
$4 (B L^{2} + C M^{2}) = 5 (A C^{2} + A B^{2})$
$B L^{2} + C M^{2} = (\frac{5}{4}) \times (A C^{2} + A B^{2})$
$B L^{2} + C M^{2} = (\frac{5}{4}) B C^{2}$
$k = (\frac{5}{4})$

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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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Δ A B C

A

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

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