Justify triangle congruence

Below are

△ A B C

and

△ D E F

. We assume that

A B = D E

B C = E F

, and

m ∠ B = m ∠ E

Here is a rough outline of a proof that

△ A B C ? △ D E F

We can map $△ A B C$ ? using a sequence of rigid transformations so that $A^{'} = D$ ? and $B^{'} = E$ ?.
If $C^{'}$ ? and $F$ ? are on the same side of $\overset{?}{D E}$ ?, then $C^{'} = F$ ?.
If $C^{'}$ ? and $F$ ? are on opposite sides of $\overset{?}{D E}$ ?, then we reflect $△ A^{'} B^{'} C^{'}$ ? across $\overset{?}{D E}$ ? and then $C^{″} = F$ ?, $A^{″} = D$ ? and $B^{″} = E$ ?.

Triangle D E F with point A prime mapped onto point D. B prime is mapped onto point E. C double prime is mapped onto point F. A dashed line extends through points D and E in both directions. The triangle is reflected across the line where A double prime is mapped onto A prime and D, B double prime is mapped onto B prime and E. C prime is equal distance away from point F on the other side of the dashed line.

What is the justification that $C^{'} = F$ ? in step 2?

(Choice A) ?
$C^{'}$ ? and $F$ ? are the same distance from $E$ ? along the same ray.
(Choice B) ?
Both $C^{'}$ ? and $F$ ? lie on intersection points of circles centered at $D$ ? and $E$ ? with radii $D F$ ? and $E F$ ?, respectively. There are two such possible points, one on each side of $\overset{?}{D E}$ ?.
(Choice C) ?
$C^{'}$ ? and $F$ ? are at the intersection of the same pair of rays.

High school geometry