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