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

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Solution

Let ABC be a right angle triangle whose angle $A$ is right angle

Let L and M be the middle point side $A C$ and $A B$ respectively.

In $Δ A C M$ ,

By Pythagoras theorem ,

$[{(Hypotenuse)}^{2} = {(B a s e)}^{2} + {(H e i g h t)}^{2}$

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

$= A C^{2} + {(\frac{1}{2} A B)}^{2}$

$C M^{2} = A C^{2} + \frac{1}{4} A B^{2}$ $. . . . . (1)$

In $Δ A B L$ ,

By Pythagoras theorem ,

${(Hypotenuse)}^{2} = {(B a s e)}^{2} + {(H e i g h t)}^{2}$

$B L^{2} = A B^{2} + A L^{2}$

$= A B^{2} + {(\frac{1}{2} A C)}^{2}$

$B L^{2} = A B^{2} + \frac{1}{4} A C^{2}$ $. . . . . (2)$

Adding equation (1) and (2) to,

$C M^{2} + B L^{2} = A C^{2} + \frac{1}{4} A B^{2} + A B^{2} + \frac{1}{4} A C^{2}$

$C M^{2} + B L^{2} = (A C^{2} + A B^{2}) + (\frac{1}{4} A B^{2} + \frac{1}{4} A C^{2})$

$C M^{2} + B L^{2} = B C^{2} + \frac{1}{4} (A B^{2} + A C^{2})$

Where $A C^{2} + A B^{2}$ $= B C^{2}$

$C M^{2} + B L^{2} = B C^{2} + \frac{1}{4} B C^{2}$

$C M^{2} + B L^{2} = \frac{5 B C^{2}}{4}$

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

Hence proved.

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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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