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Given: triangles ABC and DEF such that B = E, C = F and BC = EF.
To prove: ΔABC ≅ DEF
Proof:
Case I: If AB = DE then in
ΔABC and ΔDEF, AB = DE [by supposition] BC = EF [given] and B = E [given]
Thus, ΔABC ≅ ΔDEF [SAS criterion]
Case II: If AB < DE Take a point G on ED such that EG = AB.
Join GF. In ABC and GEF, we have AB = GE [by supposition] BC = EF [given] B = E [given] Thus, ABC GEF [SAS criterion]
ACB = GFE [corresponding parts of congruent triangles are equal]
But ACB = DFE [given] GFE =DFE,
This is only possible when FG coincides with FD or G coincides with D.
AB must be equal to DE and hence, ΔABC ≅ ΔDEF (by SAS)
Case III: If AB > ED
With a similar argument (as in case II), we may conclude that
Δ ABC ≅ ΔDEF (by SAS)
Thus,ΔABC ≅ ΔDEF.
Answer:
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