# L and M are the mid-points of AB and BC respectively of ΔABC, right-angled at B. Prove that 4LC2 = AB2 + 4BC2 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 + 4(BC)2 = 4(LC)2

Hence Proved

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