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Question
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL2 + CM2) = 5 BC2
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Solution
Given: Δ ABC right angled at A, i.e., A = 90, where BL and CM are the medians.
To Prove: 4(BL2 + CM2) = 5BC2
Proof:
Since BL is the median,
AL = CL = `1/2` AC ...(1)
Similarly, CM is the median
AM = MB = `1/2`AB ...(2)
∴ by Pythagoras theorem,
(Hypotenuse)2 = (Height)2 + (Base)2 ...(3)
In ΔBAC,
(BC)2 = (AB)2 + (AC)2 ...(4)
∠BAC = 90°
In ΔBAL,
(BL)2 = AB2 + AL2 ...(From 1)
BL2 = AB2 + `(("AC")/2)^2`
BL2 = AB2 + `("AC"^2/4)`
Multiply both sides by 4,
4BL2 = 4AB2 + AC2 ...(5)
In ΔMAC,
CM2 = AM2 + AC2 ...(From 2)
CM2 = `(("AB")/2)^2` + AC2
CM2 = `("AB")^2/4` + AC2
Multiply both sides by 4,
4CM2 = AB2 + 4AC2 ...(6)
Adding 5 and 6,
4BL2 + 4CM2 = (4AB2 + AC2) + (AB2 + 4AC2)
4(BL2 + CM2) = 5AB2 + 5AC2
∴ 4(BL2 + CM2) = 5BC2
Hence, proved.
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