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Question
In the given figure, BL and CM are medians of a ∆ABC right-angled at A. Prove that 4 (BL2 + CM2) = 5 BC2.
Solution
To prove: 4(BL2 + CM2 ) = 5BC2
Proof: In ΔCAB,
Applying Pythagoras theorem,
AB2 + AC2 = BC2 .....(1)
In ΔABL,
AL2 + AB2 = BL2
⇒ `("AC"/2)^2 + "AB"^2 = "BL"^2`
⇒ `"AC"^2 + 4"AB"^2 = 4"BL"^2` .....(2)
In ΔCAM,
CA2 + MA2 = CM2
⇒ `(("BA")/2)^2`+ CA2 = CM2
⇒ BA2 + 4CA2 = 4CM2 .....(3)
Adding (2) and (3)
AC2 + 4AB2 + BA2 + 4CA2 = 4BL2 + 4CM2
⇒ 5AC2 + 5AB2 = 4(BL2 + CM2)
⇒ 5(AC2 + AB2) = 4(BL2 + CM2)
⇒ 5(BC2) = 4(BL2 + CM2) ....( From 1 )
Hence Proved.
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