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In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:4(BL2 + CM2) = 5 BC2 - Geometry Mathematics 2

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प्रश्न

In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL+ CM2) = 5 BC2

योग
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उत्तर

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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अध्याय 2: Pythagoras Theorem - Problem Set 2 [पृष्ठ ४५]

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बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 2 Pythagoras Theorem
Problem Set 2 | Q 11 | पृष्ठ ४५

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