Selection tests for the Balkan Mathematical Olympiad 2013

Let $F^{\prime }$ be the intersection point of the tangents to circle ${\omega }_B$ at $B_1$ and $D$.



Let $G$ be the intersection point of lines $B_1{A}_1$ and $D{C}_1$.

Consider the six cyclic points $C_1,C_1,B_1,A_1,A_1,D$. Because the tangent lines to ${\omega }_B$ at $C_1$ and $A_1$ intersect at $B$, lines $C_1{B}_1$ and $A_{1}D$ intersect at $E$, and lines $B_1{A}_1$ and $D{C}_1$ intersect at $G$, from Pascal theorem, the three points $B$, $E$, and $G$ are collinear.

Consider the six cyclic points $C_1,B_1,B_1,A_1,D,D$. Because lines $C_1{B}_1$ and $A_{1}D$ intersect at $E$, the tangent lines to ${\omega }_B$ at $B_1$ and $D$ intersect at $F^{\prime }$, and lines $B_1{A}_1$ and $D{C}_1$ intersect at $G$, from Pascal theorem, the three points $E$, $F^{\prime }$, and $G$ are collinear.

We deduce that $B$, $E$, and $F^{\prime }$ are collinear and therefore $F^{\prime }=F$. This proves that $FD$ is tangent to ${\omega }_B$ at $D$.