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Answer :
Given: In \(\triangle{ABC}\) and \(\triangle{DEF}\), AB = DE, AB || DE, BC = EF and BC || EF
(i) Now, in quadrilateral ABED,
AB = DE and AB || DE ...(Given)
Since, a pair of opposite sides is equal and parallel
Therefore, ABED is a parallelogram.
(ii) In quadrilateral BEFC,
BC = EF and BC || EF ...(Given)
Since, a pair of opposite sides is equal and parallel
Therefore, BEFC is a parallelogram.
(iii)Since, ABED is a parallelogram,
AD || BE and AD = BE ...(i)
Also, BEFC in a parallelogram,
CF || BE and CF = BE ...(ii)
Thus, from Eq. (i) and (ii), we get,
AD || CF and AD = CF ...(From part (iii))
Therefore, ACFD is a parallelogram.
(v)Since, ACFD is a parallelogram.
we get, AC = DF and AC || DF
(vi)Now, in \(\triangle{ABC}\) and \(\triangle{DEF}\),
AB = DE ...(Given)
BC = EF ...(Given)
and AC = DF ...(From part (v))
Therefore, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\) ...(By SSS test)