Belarusian Mathematical Olympiad

Note that the triangle $UDV$ is a right-angled triangle regardless of the position of the point $D$ since
$$\angle UDV = \angle UDC' + \angle VDC' = \frac{1}{2} \angle ADC + \frac{1}{2} \angle BDC = \frac{1}{2} \cdot 180^\circ = 90^\circ.$$
So $S$ is the center of the circumcircle of the triangle $UDV$, and the condition $SD \perp AB$ is equivalent to the tangency of this circle and the line $AB$, and, by turn, is equivalent to the equality $\angle ADU = \angle DVU$. Let the lines $UV$ and $CD$ meet at point $T$. Since $\angle ADU = \angle CDU$, the condition $\angle ADU = \angle DVU$ is equivalent to the equality $\angle UDT' = \angle UVD$, i.e. is equivalent to the similarity $\triangle UTD \sim \triangle UDV$; this similarity is equivalent to the equality $\angle UTD = 90^\circ$. Thus, $SD \perp AB$ is equivalent to $UV \perp CD$.

Note that $U$ and $V$ are the centers of the circles $\omega_A$ and $\omega_B$ of the triangles $BDC$ and $ADC$, respectively. Let the circles $\omega_A$ and $\omega_B$ touch the side $CD$ at points $T_A$ and $T_B$, respectively. It is easy to see that the condition $UV \perp CD$ is equivalent to $T_A = T_B = T$ that is equivalent to $DT_A = DT_B$.
Let the circle $\omega_A$ touch the lines $AB$ and $BC$ at points $K$ and $L$, respectively. Using the equality of the tangents to the circle $\omega_A$, we get