A cyclic quadrilateral ABCD is such that AB = BC, AD = DC and AC and BD intersect at O. If ∠CAD = 46°, then measure of ∠AOB is equal to:

∠DCA = 46°
∠DAB + ∠DCB = 180°
46° + β + 46° + β = 180°
β = 44°
∠BCA = ∠ADB = 44°
∠DOC = ∠ADO + ∠OAD
(By exterior angle theorem)
∠DOC = 44° + 46° = 90°
Note:
In $$\square $$ ABCD
AB = BC (Given)
AD = DC (Given)

Let ∠DAC = θ
So, ∠DCA = θ (angle opposite to equal side).
∠DBC = θ (angle made by same arc DC).
∠DBA = θ (angle made by same arc AD).
Similarly,
∠ADB = ∠BDC = β
So, we can say that in a cyclic quadrilateral. If pair of adjacent sides are equal then the diagonal made by the vertex situated between equals sides is angle bisector.
In ΔAOB & ΔBOC
∠OAB = ∠OCB = β
AB = BC
∠OBA = ∠OBC = θ
ΔAOB ≅ ΔBOC (By ASA)
∴ AO = OC
Hence we can say that diagonal BD bisect diagonal AC into two equal parts.