Playtest 1

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 $\overset{\leftrightarrow}{D E}$ ‍, then $C^{'} = F$ ‍.
If $C^{'}$ ‍ and $F$ ‍ are on opposite sides of $\overset{\leftrightarrow}{D E}$ ‍, then we reflect $△ A^{'} B^{'} C^{'}$ ‍ across $\overset{\leftrightarrow}{D E}$ ‍ and then $C^{″} = F$ ‍, $A^{″} = D$ ‍ and $B^{″} = E$ ‍.

What fact can we use to justify step 1?

Math playtesting