Question

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

Answer 1

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.