Question

$ABC$ is a triangle, right-angled at $B$. $M$ is point on $BC$. Prove that :
$A{M}^2+B{C}^2=A{C}^2+B{M}^2$.

Answer 1

In $△ABC$,

${AC}^2={AB}^2+{BC}^2$ ---(1)

From $△ABM$,

${AB}^2={AM}^2-{BM}^2$ ---(2)

Substituting (2) in (1),

${AC}^2={AM}^2-{BM}^2+{BC}^2{AC}^2+{BM}^2={AM}^2+{BC}^2$

Hence proved.