Proof:  First, construct a congruent copy of triangle ABC that shares a side with DEF. 

• Construct a point C’ so that angle C’DE = angle CAB and also DC’ = AC. 
• This can be done by constructing a ray with the given angle by the protractor axiom and the point at distance AC by the ruler axiom.

• Then, since AB = DE, we can say that triangle DEC’ is congruent to triangle ABC by SAS.  Thus EC’ = CB.
• But the quadrilateral DFEC’ is a kite, since we have DF = DC’ and EF = EC’. 
• But we have proved among the kite properties that the diagonal DE divides the kite into two congruent triangles, DEF and DEC’.  But the latter triangle is congruent to triangle ABC, so we have triangle DEF is congruent to triangle ABC.

QED