AUT_ABooklet_2020

Figure 2: Problem 6

Solution:

We denote the circumcircle of the pentagon $ABCDE$ by $k$, see Figure 2. By assumption, the triangle $ABD$ is isosceles, which implies that the tangent $t_B$ to $k$ in $B$ is parallel to $AD$.

We apply Pascal's theorem to the inscribed hexagon $BEDACB$: The intersection point of the opposite sides $BE$ and $AC$ is $P$, the intersection point of the opposite sides $ED$ and $CB$ is $Q$, and the intersection point of the parallel opposite sides $BB$ (i.e., $t_B$) and $DA$ is the point at infinity corresponding to direction $AD$. Therefore, $PQ$ is parallel to $AD$.