Because $ ABCD$ is cyclic, $ \angle{BCD}+\angle{BAD}= 180^{o}$, and then a pair of angles taken from the set $ \angle{BOA},\angle{BOC},\angle{COD},\angle{DOA}$ must add up to $ 180^{o}$. Then as all these four angles together form a $ 360^{o}$ angle and have the same vertex we get 2 cases: 1. The two angles share a ray, wolog $ \angle{BOA}+\angle{BOC}= 180^{o}$. But then, $ O$ lies on $ AC$ so $ AC$ is a diameter hence $ \angle{ABC}=\angle{CDA}= 90^{o}$. then, again by the condition of the problem we must have one of angles $ \angle{BOA},\angle{BOC},\angle{COD},\angle{DOA}$ equal to $ 90^{o}$ so wolog $ \angle{COD}= 90^{o}$ then $ \angle{AOD}= 90^{o}$ so $ CD = AD$ and $ \angle{OAD}=\angle{OCD}= 45^{o}$. Then if $ \angle{BCO}= x$ then $ \angle{BAC}= 90^{o}-x$ as $ \angle{ABC}= 90^{o}$, then we get $ \angle{BAD}= 135^{o}-x$ and $ \angle{BCD}= 45^{o}+x$ while $ \angle{BOC}= 180^{o}-2x$ and $ \angle{BOA}= 2x$. Then we get $ 2x = 45^{o}+x$ giving $ x = 45^{o}$ or $ 2x = 135^{o}-x$ also giving $ x = 45^{o}$. Then $ \angle{BCD}=\angle{BAD}= 90^{o}$ therefore $ \angle{BOC}=\angle{BOA}= 90^{o}$ so $ BC = CD = AD = BA$ as triangles $ BOC, COD, DOA, AOB$ are congruent so as all angles of $ ABCD$ are $ 90^{o}$ $ ABCD$ is a square. 2. The two angles do not share a ray, wolog $ \angle{BOA}+\angle{COD}= 180^{o}$ then $ \angle{BDA}+\angle{CAD}= 90^{o}$ therefore $ BD$ and $ AC$ intersect at a $ 90^{o}$ angle giving $ ABCD$ is an isosceles trapezoid. Therefore, wolog $ BA = CD$ and then $ BC$ is parallel to $ AD$, but then $ \angle{ABC}=\angle{BCD}$ and $ \angle{BAD}=\angle{CDA}$. If all four angles are equal we get $ ABCD$ as a rectangle and complete the prff in the same way as in case 1. Otherwise, by the condition of the problem, this means that either $ \angle{BOA}=\angle{COD}$ and $ \angle{BOC}=\angle{AOD}$ resulting in each of the four angles being $ 90^{o}$ as $ \angle{BOA}+\angle{COD}= 180^{o}$ or $ \angle{BOA}=\angle{BOC}$ and $ \angle{AOD}=\angle{COD}$. But then $ AB = BC$ and $ AD = CD$ so $ AB = BC = CD = DA$ so $ ABCD$ is a rhombus and because it is inscribed in a circle, it is a square. QED