62nd Ukrainian National Mathematical Olympiad

Let $P$ be the point of intersection of the rhombus diagonals, $O_1$ and $O_2$ be the centers of the described circles $\Delta ATC$ and $\Delta BTD$ respectively (Fig. 1). Hence, the following equalities are true: $$\angle A{O}_{1}C=2\angle ATC=2\cdot (180^{\circ }-\angle BTD)=\angle B{O}_{2}D,$$ so the isosceles triangles $A{O}_{1}C$ and $B{O}_{2}D$ are similar. Then their altitudes are proportional to their sides, i.e. $\frac{P{O}_1}{P{O}_2}=\frac{O_{1}A}{O_{2}B}=\frac{O_{1}T}{O_{2}T}$. Then $PT$ is a bisector (internal or external) of $\angle O_{1}P{O}_2$, exactly what we needed to prove.