In $\Delta ABC$ and $\Delta PQR$,if $\angle B = \angle P = 90^{\circ}$,$AC = RQ$,and $AB = RP$,then $\Delta ABC \cong \Delta \ldots \ldots \ldots$

In triangles $ABC$ and $DEF$,$AB = FD$ and $\angle A = \angle D$. The two triangles will be congruent by the $SAS$ axiom if:

In $\Delta ABC$ and $\Delta PQR$,$AB = PR$,$BC = RQ$ and $\angle B = \angle R$,then $\Delta ABC \cong \Delta \ldots$

In the given figure,$PQ = PR$ and $\angle Q = \angle R$. Prove that $\triangle PQS \cong \triangle PRT$.

$\Delta ABC$ is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively. Show that these altitudes are equal.

In the given figure,$PN$ and $QM$ are both perpendicular to line segment $PQ$. Also,$X$ is the midpoint of $PQ$ as well as $MN$. Prove that $\triangle PNX \cong \triangle QMX$.