On page 97 of the textbook, when proving that "two triangles with proportional sides and equal included angles are similar" (as shown in the figure, it is known that DE/AB = DF/AC (AB > DE), ∠A=∠D, and verification: △ABC∠△DEF), the mathematical idea of transformation is used, and the unknowns are added with auxiliary lines. Please use the above method to prove this theorem.

It is proved that the intercept AG=DE on AB is GH∨BC.

∴△AGH∽△ABC

∴AG/AB=AH/AC

∫DE/AB = DF/AC，AG=DE

∴AH=DF

∠∠A =∠D

∴△AGH≌△DEF(SAS)

∴△ABC∽△DEF

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