In $Δ$ ABC, $∠$ BAC $= 90^{o}$ , seg BL and seg CM are medians of $Δ$ ABC. Then prove that: $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .

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Solution

Given, $Δ A B C$ is right-angled at $A$ i.e. $∠ A = 90^{0}$ . where $B L, C M$ are medians.
Since $B L$ is median, $A L = C L = \frac{1}{2} A C$
Similary, $C M$ is median. $A M = M B = \frac{1}{2} A B$ ------(2)
Apply pythagoras theoram in $Δ B A C$ ,
$B C^{2} = A B^{2} + A C^{2}$ ........(i)
In $Δ B A L, B L^{2} = A B^{2} + A L^{2} \Rightarrow B L^{2} = A B^{2} + \frac{A C^{2}}{4}$
$\Rightarrow 4 B L^{2} = 4 A B^{2} + A C^{2}$ ..........(ii)
In $Δ M A C, C M^{2} = A M^{2} + A C^{2} \Rightarrow C M^{2} = \frac{A B^{2}}{4} + A C^{2}$
$\Rightarrow 4 C M^{2} = A B^{2} + 4 A C^{2}$ .........(iii)
Adding (ii) and (iii)
$\Rightarrow 4 B L^{2} + 4 C M^{2} = 4 A B^{2} + A C^{2} + A B^{2} + 4 A C^{2}$
$\Rightarrow 4 (B L^{2} + C M^{2}) = 5 B C^{2}$
$∴$ Hence proved.

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