Belarusian Mathematical Olympiad

Note that the triangle $UDV$ is a right-angled triangle regardless of the position of the point $D$ since $$\angle UDV=\angle UD{C}^{\prime }+\angle VD{C}^{\prime }=\frac{1}{2}\angle ADC+\frac{1}{2}\angle BDC=\frac{1}{2}\cdot 180^{\circ }=90^{\circ }.$$ 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 $\angle UD{T}^{\prime }=\angle UVD$, i.e. is equivalent to the similarity $△UTD\sim △UDV$; 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 $D{T}_A=D{T}_B$. 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