Q. 185.0( 4 Votes )

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

Answer :


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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