6.5 Pythagoras Theorem

In right $\Delta ABC$: $BC^2 = AB^2 + AC^2$ (1).

In right $\Delta BAL$ (L is mid-point of AC): $BL^2 = AB^2 + AL^2 = AB^2 + (AC/2)^2$.

$4BL^2 = 4AB^2 + AC^2$ (2).

In right $\Delta CAM$ (M is mid-point of AB): $CM^2 = AC^2 + AM^2 = AC^2 + (AB/2)^2$.

$4CM^2 = 4AC^2 + AB^2$ (3).

Add (2) and (3): $4(BL^2 + CM^2) = (4AB^2 + AB^2) + (AC^2 + 4AC^2) = 5AB^2 + 5AC^2$.

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