Prove that in a right angle …

According to question, we are given a triangle ABC in which AC² = AB² + BC².
We need to prove that {tex}\angle B=90^{\circ}{/tex}
To start with, we construct a {tex}\Delta \mathrm{PQR}{/tex} right angled at Q such that PQ = AB and QR = BC.
 
Now, we observe that from {tex}\Delta{/tex}PQR, we have :
PR² = PQ² + QR² (Pythagoras Theorem, as {tex}\angle Q=90^{\circ}{/tex})
PR² = AB² + BC² ..................(i) (By construction)
But, AC² = AB² + BC² ...............(ii) (given)
Therefore, AC = PR ................(iii)  [From (i) and (ii)]
Now, in {tex}\Delta{/tex}ABC and {tex}\Delta{/tex}PQR,
AB = PQ (By construction)
BC = QR (By construction)
AC = PR [proved in (iii) above]
Therefore, {tex}\Delta ABC \cong \Delta PQR{/tex} (SSS congruence)
Hence, {tex}\angle{/tex}B = {tex}\angle{/tex}Q
But {tex}\angle{/tex}Q = {tex}{90^ \circ }{/tex} (By Construction)
Therefore, {tex}\angle{/tex}B = {tex}{90^ \circ }{/tex}