Given: abcd is a trapezoid, m∠1=m∠2, m∠3=m∠4, ab = cd, ac ∩ bd = o
prove: m∠aod = m∠bcd

Since AB = CD the trapezoid is isosceles, which means that ∡A = ∡D

Therefore also ∡2 = ∡3 (they are half of the congruent angles)

For the properties of parallel lines (BD and AD) crossed by a transversal (BD) we have ∡3 = ∡CBD.

Now consider triangles AOD and BCD:
∡OAD (2) = ∡ADO (3) = ∡CBD (3) = ∡CDB (4)

The sum of the angles of a triangle must be 180°, therefore:
∡AOD = 180 - ∡2 - ∡3
∡BCD = 180 - ∡3 - ∡4

∡AOD = ∡BCD because their measure is the difference of congruent angles.