In $\Delta ABC$ and $\Delta XYZ$,$\angle A = \angle X$,$\angle C = \angle Z$ and $AB = XY$,then $\Delta ABC \cong \Delta \ldots \ldots \ldots$

$CDE$ is an equilateral triangle formed on side $CD$ of a square $ABCD$ (see figure). Show that $\triangle ADE \cong \triangle BCE$.

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

$AB$ is a line segment. $P$ and $Q$ are points on opposite sides of $AB$ such that each of them is equidistant from the points $A$ and $B$ (see figure). Show that the line $PQ$ is the perpendicular bisector of $AB$.

$ABCD$ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see figure). Prove that:
$(i)$ $\Delta ABD \cong \Delta BAC$
$(ii)$ $BD = AC$
$(iii)$ $\angle ABD = \angle BAC$

In an isosceles triangle $ABC$ with $AB = AC$,$D$ and $E$ are points on $BC$ such that $BE = CD$ (see Fig). Show that $AD = AE$.