Question

$L$ and $M$ are the mid-point of $AB$ and $BC$ respectively of $△ABC$ right-angled at $B$. Prove that $4L{C}^2=A{B}^2+4B{C}^2$.

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)${}^2$

$⟹(LB{)}^2+(BC{)}^2=(LC{)}^2$
$⟹(\frac{AB}{2}{)}^2+(BC{)}^2=(LC{)}^2$
$⟹(AB{)}^2+4(BC{)}^2=4(LC{)}^2$