Given:    Trapezoid ABCD,  
          AC = DB
To prove: AB = DC


AD ∥ BC                          definition of a trapezoid
∠DAC = ∠BCA, ∠ADB = ∠DBC       alternate interior angles
ΔADE ∽ ΔCBE                     two angles equal in each  
AE/EC = DE/EB                   corresponding sides of similar 
                                        triangles are proportional
AE/EC+1 = DE/EB+1               adding the same quantity, 1, to both sides
AE/EC+EC/EC = DE/EB+EB/EB       replacing 1 by EC/EC and by EB/EB
(AE+EC)/EC = (DE+EB)/EB         adding fractions 
AE+EC = AC,  DE+EB = DB         a whole is the sum of its parts           
AC/EC = DB/EB                   Substitution of equals for equals
AC = DB                         Given
AC/EC = AC/EB                   Substitution of equals for equals
AC*EB = AC*EC                   Cross-multiplication or product of
                                extremes equal product of means.
EB = EC                         Dividing both sides by AC                             
ΔEBC is isosceles               two sides equal
∠ECB = ∠EBC                    base angles of an isosceles triangle
AC = DB                         given
BC = BC                         identity
∠ACB = ∠DBC                    Same as ∠ECB = ∠EBC 
ΔABC ≅ ΔDCB                    SAS   
AB = DC                         Corresponding parts of congruent triangles

Edwin