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

Step 1: Convert the given data into the figure.
BL and CM are medians of ∆ABC in which ∠A = 90°
Step 2: Make a note of all the right triangles in the figure.
NOTE: Right triangles
\triangle ABC
\triangle ABL
\triangle AMC
Step 3: Apply the Pythagoras theorem to all the right angle triangles in the figures.
EXPLANATION:
\triangle ABC
BC^2 = AB^2 + AC^2 ............................(1)
\triangle ABL
BL^2 = AB^2 + AL^2 ...........................(2)
\triangle AMC
CM^2 = AM^2 + AC^2 .....................................(3)
Step 4: Use the given hints in the question to prove the required equation.
EXAMPLE: From equation (2)
BL^2 = AB^2 + AL^2
BL^2 = AB^2 + (\frac{AC}{2})^2 (∵ BL is median, L is the midpoint of AC)
4BL^2 = 4AB^2 + AC^2 .......................(4)
From equation (3)
CM^2 = (\frac{AB}{2})^2 + AC^2 (∵ CM is median, M is the midpoint of AB)
4CM^2 = AB^2 + 4AC^2 ...........................(5)
Step 5: Add equation (4) and (5)
4CM^2 + 4BL^2 = 4AB^2 + AC^2 + AB^2 + 4AC^2
4(CM^2 + BL^2) = 5AB^2 + 5AC^2
4(CM^2 + BL^2) = 5(BC)^2 [Since equation (1)]
Hence proved