A student solved the following problem and made an error:

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A student solved the following problem and made an error:

Geometry
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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1 Attachment
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 2 minus x sub 1 squared plus quantity y sub 2 minus y sub 1 squared, then d equals square root of quantity 0 minus 2 all squared plus quantity 2 minus 4 all squared, then d equals square root of negative 2 squared plus negative 2 squared, then d equals square root of 4 plus 4, then d equals square root of 8 segment AB = 2.83 Line 4 Length of segment FD. D (2, 0) F (4, 2) d equals square root of quantity x sub 2 minus x sub 1 squared plus quantity y sub 2 minus y sub squared then d equals square root of quantity 2 minus 4 all squared plus quantity 0 minus 2 all squared, then d equals square root of negative 2 squared plus negative 2 squared, then d equals square root of 4 plus 4, then d equals 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? Line 3 Line 5 Line 1 Line 2

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Other answers:

Woah
I agree @sunleaf01
@SomeoneHelpPlz What is the student trying to prove?

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