, 2016

Let us use Pascal's Theorem for the points $A, B, C$ and $D$ on the circle. Here we use the points $B$ and $D$ twice and denote their copies by $B_1$ and $D_1$. We have $AB \cap CD = E$, $BB_1 \cap DD_1 = F$ and the lines $AD_1$ and $B_1C$ either intersect or are parallel.

If $AD_1$ and $B_1C$ intersect, denote this intersection by $G$. The Pascal's Theorem implies that $E, F$ and $G$ are collinear. We now wish to apply Ceva's Theorem for the triangle $AEG$ and the points $B, F$ and $D$ to show that $AF, BG$ and $DE$ are concurrent. The lines $BG$ and $DE$ intersect at $C$, so it suffices to show that
$$\frac{AB \cdot EF \cdot GD}{BE \cdot FG \cdot DA} = 1. \qquad (4)$$
Since $AB \parallel DF$ the similarity of the triangles $AEG$ and $DFG$ implies that $\frac{EF}{FG} = \frac{DA}{GD}$. The point $B$ is the midpoint of the segment $AE$, so we see that (4) does indeed hold. The lines $AF, BG$ and $DE$ intersect at $C$ and the points $A, C$ and $F$ are collinear.

If $AD_1$ and $B_1C$ are parallel then by Pascal's Theorem the line $EF$ is parallel to both of them as well. So, $AEFD$ is a parallelogram and the point $C$ is the midpoint of the diagonal $DE$. The other diagonal also passes through $C$, so the points $A, C$ and $F$ are collinear.

Let $T$ be the intersection of the lines $AC$ and $BD$, and let the point $T'$ be the intersection of $AE$ and $BD$. We wish to show that $T = T'$. Since $AE$ is a symmedian we have
$$\frac{|BT'|}{|T'D|} = \frac{|AB|^2}{|AD|^2}.$$
Since $T$ and $T'$ both lie on the segment $BD$, it suffices to show that for $T$ we also have
$$\frac{|BT|}{|TD|} = \frac{|AB|^2}{|AD|^2}. \qquad (5)$$
Remark: This part of the reasoning can also be done using Menelaus' Theorem and Power-of-a-Point Theorem instead of the Symmedian Theorem.

By Menelaus' Theorem for the triangle $BED$ and the collinear points $A, T$ and $C$ we have
$$\frac{|BT|}{|TD|} = \frac{|DC|}{|CE|} = \frac{|EA|}{|AB|} = 1.$$
Since $|EA| = 2|AB|$, we get
$$\frac{|BT|}{|TD|} = \frac{|CE|}{2|DC|}.$$
The Power-of-a-Point Theorem for the point $E$ and the circle $K$ states that $|EC| \cdot |ED| = |EB| \cdot |EA| = 2|AB|^2$. So,
$$\frac{|BT|}{|TD|} = \frac{2|AB|^2}{2|DC| \cdot |ED|} = \frac{|AB|^2}{|DC| \cdot |ED|}.$$
Since $AB \parallel DF$ and $DF$ is tangent to $K$, $D$ must be the midpoint of the arc $\widehat{BA}$. The triangle $ABD$ is therefore isosceles with the apex at $D$ and $|AD| = |BD|$. It follows that
$$\asymp DCB = \pi - \asymp BAD = \pi - \asymp DBA = \asymp EBD.$$
This implies that the triangles $DBC$ and $DEB$ are similar and
$$\frac{|DC|}{|DB|} = \frac{|DB|}{|DE|} \Rightarrow |DC| \cdot |ED| = |DB|^2 = |AD|^2.$$
Thus,
$$\frac{|BT|}{|TD|} = \frac{|AB|^2}{|AD|^2}.$$
Therefore, $T = T'$, so the points $A, C$ and $F$ are collinear.