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In the given figure, BL and CM are medians of a ∆ABC right-angled at A. Prove that 4 (BL^{2} + CM^{2}) = 5 BC^{2}.

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#### Solution

To prove: 4(BL^{2 }+ CM^{2} ) = 5BC^{2}

Proof: In ΔCAB,

Applying Pythagoras theorem,

AB^{2 }+ AC^{2 }= BC^{2 } .....(1)

In ΔABL,

AL^{2 }+ AB^{2} = BL^{2}

⇒ `("AC"/2)^2 + "AB"^2 = "BL"^2`

⇒ `"AC"^2 + 4"AB"^2 = 4"BL"^2` .....(2)

In ΔCAM,

CA^{2 }+ MA^{2} = CM^{2}⇒ `(("BA")/2)^2`+ CA^{2 }= CM^{2}

⇒ BA^{2 }+ 4CA^{2} = 4CM^{2} .....(3)

Adding (2) and (3)

AC^{2} + 4AB^{2} + BA^{2} + 4CA^{2} = 4BL^{2} + 4CM^{2}

⇒ 5AC^{2} + 5AB^{2} = 4(BL^{2} + CM^{2})

⇒ 5(AC^{2} + AB^{2}) = 4(BL^{2} + CM^{2})

⇒ 5(BC^{2}) = 4(BL^{2} + CM^{2}) ....( From 1 )

Hence Proved.

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