If $\Delta ABC \cong \Delta RPQ,$ then which side of $\Delta PQR$ is equal to side $AB$ of $\Delta ABC$?

In right triangle $ABC$,right-angled at $C$,$M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$ (see Fig). Show that:
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$

It is given that $\triangle ABC \cong \triangle FDE$. If $AB = 5 \, cm$,$\angle B = 40^{\circ}$,and $\angle A = 80^{\circ}$,then which of the following is true?

If $\Delta ABC \cong \Delta YZX,$ then in $\Delta XYZ,$ which side is equal to side $AB$?

$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.

For the correspondence $ABC \leftrightarrow PRQ$ between $\Delta ABC$ and $\Delta PQR,$ the side $\ldots$ corresponds to $AB.$