Question

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

Answer 1

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