Given: ABC is a right triangle right angled at B
and L and M are the mid-points of AB and BC respectively.
⇒ AL = LB and BM = MC
In ∆LBC, using Pythagoras theorem we have,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (LB)2 + (BC)2 = (LC)2
⇒ (AB/2)2 + (BC)2 = (LC)2
⇒ (AB)2 + 4(BC)2 = 4(LC)2
Hence Proved