In \(\triangle ABC\), \(\angle BAC = 90^\circ\), seg BL and seg CM are medians of \(\triangle ABC\). Then prove that: \(4(BL^2 + CM^2) = 5\,BC^2\)
Figure: L is the midpoint of side AC and M is the midpoint of side AB.
Answer: Pythagoras theorem in \(\triangle BAL\), \(\triangle CAM\) and \(\triangle BAC\) gives the result.
L is the midpoint of AC, so AL = \(\frac{AC}{2}\); M is the midpoint of AB, so AM = \(\frac{AB}{2}\).
In right-angled \(\triangle BAL\):
\[BL^2 = AB^2 + \frac{AC^2}{4}\]In right-angled \(\triangle CAM\):
\[CM^2 = AC^2 + \frac{AB^2}{4}\]Adding and multiplying by 4:
\[4(BL^2 + CM^2) = 5AB^2 + 5AC^2\] \[= 5(AB^2 + AC^2) = 5BC^2\](since \(AB^2 + AC^2 = BC^2\) in right-angled \(\triangle BAC\)). Hence proved.
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