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

(N/A) $BL$ and $CM$ are medians of $\Delta ABC$ in which $\angle A = 90^{\circ}$.
From $\Delta ABC$,by Pythagoras theorem:
$BC^2 = AB^2 + AC^2$ $...(1)$
From $\Delta ABL$,since $L$ is the midpoint of $AC$,$AL = AC/2$:
$BL^2 = AB^2 + AL^2 = AB^2 + (AC/2)^2 = AB^2 + AC^2/4$
$4BL^2 = 4AB^2 + AC^2$ $...(2)$
From $\Delta CMA$,since $M$ is the midpoint of $AB$,$AM = AB/2$:
$CM^2 = AC^2 + AM^2 = AC^2 + (AB/2)^2 = AC^2 + AB^2/4$
$4CM^2 = 4AC^2 + AB^2$ $...(3)$
Adding $(2)$ and $(3)$:
$4(BL^2 + CM^2) = 4AB^2 + AC^2 + 4AC^2 + AB^2$
$4(BL^2 + CM^2) = 5AB^2 + 5AC^2$
$4(BL^2 + CM^2) = 5(AB^2 + AC^2)$
Using equation $(1)$,$AB^2 + AC^2 = BC^2$,so:
$4(BL^2 + CM^2) = 5BC^2$.