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

We know, By Apollonius theorem

In ΔABC, if L is the midpoint of side AC, then AB² + BC² = 2BL² + 2AL ²

Given that, BL is median i.e. L is the mid-point of CA

⇒ AB² + BC² = 2BL² + 2AL²

[1]

Also, if M is the midpoint of side AB, then AC² + BC² =2CM² + 2BM²

Given that, CM is median i.e. M is the mid-point of BA

⇒ AC² + BC² = 2CM² + 2BM²

[2]

Also, In ΔABC, By Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

⇒ BC² = AC² + AB² [3]

Adding [1] and [2]

⇒ AB² + AC² + 4BC² = 4(BL² + CM²)

⇒ BC² + 4BC² = 4(BL²+ CM²) [From 3]

⇒ 5BC² = 4(BL²+ CM²)

Hence Proved.