(N/A) $1$. In $\Delta PQR$,let $PL \perp QR$,$QM \perp PR$,and $RN \perp PQ$. Given $PL = QM = RN$.
$2$. Consider $\Delta QRN$ and $\Delta RQM$. We have $\angle QNR = \angle RMQ = 90^{\circ}$,$QR = RQ$ (common side),and $RN = QM$ (given).
$3$. By $RHS$ congruence,$\Delta QRN \cong \Delta RQM$. Thus,$\angle RQN = \angle QRM$,which implies $\angle RQP = \angle QRP$.
$4$. Similarly,by considering $\Delta PQL$ and $\Delta QPM$,we can show $\angle QPR = \angle PQR$.
$5$. Since $\angle PQR = \angle QRP$ and $\angle QPR = \angle PQR$,it follows that $\angle PQR = \angle QRP = \angle QPR$.
$6$. Since all angles are equal,$\Delta PQR$ is an equilateral triangle.