Two Symmetric Triangles Are Directly Equidecomposable II

The problem #9 in a delightful collection Which Way Did the Bicycle Go? reads:

Can you cut an arbitrary triangle into pieces so that the pieces can be rotated and translated (but not flipped) so as to form the mirror image of the given triangle? It can be done in just two cuts.

My attempts at solving this problem led me a little astray. I was put back on the right track with the help from Nathan Bowler. To all intents and purposes this is indeed the intended solution, but the manner in which it's arrived at is different.

References

1. J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996, #9

Equidecomposition by Dissiection

Can you cut an arbitrary triangle into pieces so that the pieces can be rotated and translated (but not flipped) so as to form the mirror image of the given triangle? It can be done in just two cuts.

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