In the given figure,if $PQ = ST$,$QU = TR$,$PQ \perp QT$ and $ST \perp TQ$,then prove that $PR = SU$.

(N/A) Given: $PQ = ST$,$QU = TR$,$\angle PQT = 90^\circ$ and $\angle STQ = 90^\circ$.
To prove: $PR = SU$.
Proof:
Consider $\triangle PQR$ and $\triangle STU$.
$1$. $PQ = ST$ (Given)
$2$. $\angle PQR = \angle STU = 90^\circ$ (Given)
$3$. $QR = QU + UR$ and $TU = TR + UR$.
Since $QU = TR$,we have $QR = TR + UR = TU$.
Thus,$QR = TU$.
By $SAS$ congruence criterion,$\triangle PQR \cong \triangle STU$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$PR = SU$.