Question

BL and CM are medians of a△ABC, right-angle at A. Prove that
4(BL² + CM² = 5BC²).​

Answer 1

• Given

A △ABC in which BL and CM are medians and ∠A = 90°

• To Prove

4(BL² + CM²) = 5BC²

• Proof

In △BAC, ∠A = 90°

∴ BC² = AB² + AC² ------------ (i)

[ By Pythagoras theorem ]

.

In △BAL, ∠A = 90°

∴ BL² = AL² + AB²

[ By Pythagoras theorem ]

⇒ BL² = ${(\frac{1}{2}\:AC)}^{2}$ + AB²

⇒ BL² = ${(\frac{1}{4}\:AC)}^{2}$ +AB²

⇒ 4BL² = AC² + 4AB² ------------ (ii)

.

In △CAM, ∠A = 90°

∴ CM² = AM² + AC²

⇒ CM² = ${(\frac{1}{2}\:AB)}^{2}$ + AC²

⇒ CM² = ${(\frac{1}{4}\:AC)}^{2}$ + AB² + AC²

⇒ 4CM² = AB² + 4AC² ------------ (iii)

.

On adding (ii) and (iii), we get

4(BL² + CM²) = 5(AB² + AC²).

.

Hence, 4(BL² + CM²) = 5BC² [ using (i) ]

Answer 2

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$\huge\tt\bf\underline{Solution}$

A △ABC in which BL and CM are medians and ∠A = 90°

To Prove -

4(BL² + CM²) = 5BC²

In △BAC, ∠A = 90°

∴ BC² = AB² + AC² ────── (i)

[ By Pythagoras theorem ]

ㅤㅤ

In △BAL, ∠A = 90°

∴ BL² = AL² + AB²

[ By Pythagoras theorem ]

⇒ BL² = ${(\frac{1}{2}\:AC)}^{2}{(AC)}^{2}$2 + AB²

⇒ BL² =${(\frac{1}{4}AC)}^{2}(AC)$ 2 +AB²

⇒ 4BL² = AC² + 4AB² ───── (ii)

ㅤㅤ

In △CAM, ∠A = 90°

∴ CM² = AM² + AC²

⇒ CM² = ${(\frac{1}{2}\:AB)}^{2}(AB)$2 + AC²

⇒ CM² =${(\frac{1}{4}AC)}^{2}(AC)$2 + AB² + AC²

⇒ 4CM² = AB² + 4AC² ───── (iii)

ㅤㅤㅤㅤ

On adding (ii) and (iii), we get

On adding (ii) and (iii), we get4(BL² + CM²) = 5(AB² + AC²).

ㅤㅤㅤㅤ

Hence, 4(BL² + CM²) = 5BC² [ using (i) ]

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