BL and CM are medians of a triangle ABC right angled at A. Prove that $ 4(BL^2 +CM^2 )=5BC^2 $

Question

BL and CM are medians of a triangle ABC right angled at A. Prove that $ 4(BL^2 +CM^2 )=5BC^2 $

My attempt: I have found out till $ 4BL^2 =4AB^2 +(AC)^2 $. I am not getting how to move forward. Any help will be appreciated.

Any help will be appreciated.

Answer 1

You have found $4BL^2 = 4AB^2 + AC^2$. Similarly we would have $4CM^2 = 4AC^2 + AB^2$.

Adding them together, we have:

$$4(BL^2 + CM^2) = 4AB^2 + AC^2 + 4AC^2 + AB^2 = 5(AB^2 + AC^2) = 5BC^2$$

The final equality is due to another application of the Pythagorean Theorem.

Answer 2

Both $BL$ and $CM$ are medians, so use Apollonius's Theorem twice. $$2AB^2+2BC^2=4BL^2+AC^2$$ $$2BC^2+2AC^2=4CM^2+AB^2$$ Add the two equations and use the fact that (from Pyhtagorean Theorem), $$AB^2+AC^2=BC^2$$ to finally show that, $$4(BL^2 +CM^2 )=5BC^2.$$