Prove that if in a triangle the square of one side is equal to the sum of the squares of the other two sides,then the angle opposite the first side is a right angle.

(N/A) Given: $A$ triangle $ABC$ such that $AC^2 = AB^2 + BC^2$.
To prove: $\angle B = 90^{\circ}$.
Construction: Construct a $\triangle PQR$ such that $\angle Q = 90^{\circ}$,$PQ = AB$,and $QR = BC$.
Proof:
In $\triangle PQR$,by Pythagoras theorem:
$PR^2 = PQ^2 + QR^2$
Since $PQ = AB$ and $QR = BC$,we have:
$PR^2 = AB^2 + BC^2$ ...... $(1)$
Given that $AC^2 = AB^2 + BC^2$ ...... $(2)$
From $(1)$ and $(2)$,$PR^2 = AC^2$,which implies $PR = AC$.
Now,in $\triangle ABC$ and $\triangle PQR$:
$AB = PQ$ (Construction)
$BC = QR$ (Construction)
$AC = PR$ (Proved above)
Therefore,$\triangle ABC \cong \triangle PQR$ by $SSS$ congruence criterion.
Thus,$\angle B = \angle Q$ by $CPCT$.
Since $\angle Q = 90^{\circ}$ by construction,it follows that $\angle B = 90^{\circ}$.