Below are \triangle ABC△ABCtriangle, A, B, C and \triangle DEF△DEFtriangle, D, E, F. We assume that AB=DEAB=DEA, B, equals, D, E, m\angle A=m\angle Dm∠A=m∠Dm, angle, A, equals, m, angle, D, and m\angle B=m\angle Em∠B=m∠Em, angle, B, equals, m, angle, E. Here is a rough outline of a proof that \triangle ABC\cong\triangle DEF△ABC≅△DEFtriangle, A, B, C, \cong, triangle, D, E, F:
We can map \triangle ABC△ABCtriangle, A, B, C using a sequence of rigid transformations so that A'=DA
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=DA, prime, equals, D and B'=EB
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=EB, prime, equals, E. [Show drawing.]
If C'C
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C, prime and FFF are on the same side of \overleftrightarrow{DE}
DE
D, E, with, \overleftrightarrow, on top, then C'=FC
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=FC, prime, equals, F. [Show drawing.]
If C'C
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C, prime and FFF are on opposite sides of \overleftrightarrow{DE}
DE
D, E, with, \overleftrightarrow, on top, then we reflect \triangle A'B'C'△A
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B
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C
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triangle, A, prime, B, prime, C, prime across \overleftrightarrow{DE}
DE
D, E, with, \overleftrightarrow, on top and then C''=FC
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=FC, start superscript, prime, prime, end superscript, equals, F, A''=DA
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=DA, start superscript, prime, prime, end superscript, equals, D and B''=EB
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=EB, start superscript, prime, prime, end superscript, equals, E. [Show drawing.]
Answer two questions about this proof.