Given:- \\\\(ABC\\\\) is a triangle\\\\({AC}^{2} = {AB}^{2} + {BC}^{2}\\\\)To prove:- \\\\(\\angle{B} = 90°\\\\)Construction:- Construct a triangle \\\\(PQR\\\\) right angled at \\\\(Q\\\\) such that, \\\\(PQ = AB\\\\) and \\\\(QR = BC\\\\)Proof:-In \\\\(\ riangle{PQR}\\\\)\\\\({PR}^{2} = {PQ}^{2} + {QR}^{2} \\quad \\left( \ ext{By pythagoras theorem} \\right)\\\\)\\\\(\\Rightarrow {PR}^{2} = {AB}^{2} + {BC}^{2}..... \\left( 1 \\right) \\quad \\left( \\because AB = PQ \ ext{ and } QR = BC \\right)\\\\)\\\\({AC}^{2} = {AB}^{2} + {BC}^{2} ..... \\left( 2 \\right) \\quad \\left( \ ext{Given} \\right)\\\\)From equation \\\\(\\left( 1 \\right) \\& \\left( 2 \\right)\\\\), we have\\\\({AC}^{2} = {PR}^{2}\\\\)\\\\(\\Rightarrow AC = PR ..... \\left( 3 \\right)\\\\)Now, in \\\\(\ riangle{ABC}\\\\) and \\\\(\ riangle{PQR}\\\\)\\\\(AB = PQ\\\\)\\\\(BC = QR\\\\)\\\\(AC = PR \\quad \\left( \ ext{From } \\left( 3 \\right) \\right)\\\\)\\\\(\ herefore \ riangle{ABC} \\cong \ riangle{PQR} \\quad \\left( \ ext{By SSS congruency} \\right)\\\\)Therefore, by C.P.C.T.,\\\\(\\angle{B} = \\angle{Q}\\\\)\\\\(\\because \\angle{Q} = 90°\\\\)\\\\(\ herefore \\angle{B} = 90°\\\\)Hence proved.