$PL, QM$ and $RN$ are the altitudes of $\Delta PQR$. If $PL = QM = RN$,then by $RHS$ criterion of congruence,prove that $\Delta PQR$ is an equilateral triangle.

(N/A) $1$. Consider the right-angled triangles $\Delta PLQ$ and $\Delta QMP$.
$2$. In these triangles,$PL = QM$ (given),$PQ = QP$ (common side),and $\angle PLQ = \angle QMP = 90^{\circ}$ (altitudes).
$3$. By $RHS$ congruence criterion,$\Delta PLQ \cong \Delta QMP$.
$4$. By $CPCT$,$\angle PQL = \angle QPM$,which implies $\angle PQR = \angle QPR$.
$5$. Since the base angles are equal,the sides opposite to them are equal,so $PR = QR$.
$6$. Similarly,by considering $\Delta QMR$ and $\Delta RNP$,we can prove $\Delta QMR \cong \Delta RNP$,which implies $\angle QRM = \angle RPN$,leading to $PQ = QR$.
$7$. Since $PR = QR$ and $PQ = QR$,we have $PQ = QR = PR$.
$8$. Therefore,$\Delta PQR$ is an equilateral triangle.