A student solved the following problem and made an error:

Triangles ABC and DEF. Angles A and F are congruent and measure 135 degrees. The coordinates for the vertices are at A 0, 2 and B 2, 4 and C 0, 0 and D 2, 0 and E 4, 4 and F 4, 2.


Line 1 segment AC equals 2, segment DF equals 2, segment AC is congruent to segment DF
Line 2 ∠A ≅ ∠F
Line 3 Length of segment AB.
A (0, 2)
B (2, 4)
d equals square root of quantity x sub 1 minus x sub 2 squared plus quantity y sub 1 minus y sub 2 squared, then d equals the square root of quantity 0 minus 2 all squared plus quantity 2 minus 4 all squared, then d equals the square root of negative 2 squared plus negative 2 squared, then d equals the square root of 4 plus 4, then d equals the square root of 8
segment AB = 2.83
Line 4 Length of segment FD.
D (2, 0)
F (4, 2)
d equals the square root of quantity x sub 1 minus x sub 2 squared plus quantity y sub 1 minus y sub 2 squared then d equals the square root of quantity 2 minus 4 all squared plus quantity 0 minus 2 all squared, then d equals the square root of negative 2 squared plus negative 2 squared, then d equals the square root of 4 plus 4, then d equals the square root of 8
segment FD = 2.83
Line 5 segment AC is congruent to segment FD
Line 6 triangle ABC is congruent to triangle FDE by SAS

In which line did the student make the first mistake?