First, we reformulate the problem as follows: Reformulated problem wrote: Let $A$ and $B$ be a point on a circle $\Gamma$. $\Gamma_2$ is a circle passing through $A$, meeting $\Gamma$ for the second time at $F$ on the minor arc $AB$, and such that $B$ is outside of it. Let the tangent lines from $B$ to $\Gamma_2$ intersect $\Gamma$ at $Z_1$ and $Z_2$, respectively. Let $C$ be another point on $\Gamma$ not on minor arc $AB$. $\Gamma_1$ is a circle passing through $A$ and is tangent to $CZ_1$ and $CZ_2$, such that it meets $\Gamma$ for the second time at $E$ on the minor arc $AC$. Let $D$ be the second intersection point of $\Gamma_1$ and $\Gamma_2$. We are going to prove that $F, D, C$ (and similarly $E, D, B$) are collinear. Perform inversion with center $A$. The problem is again reformulated, as follows. Reformulated problem after inversion at A wrote: Let $A$ be a point and $\ell$ be a line. $B, Z_1, Z_2, C$ are points on $\ell$. $\Gamma_2$ is the common tangent line of $\odot(AZ_1B)$ and $\odot(AZ_2B)$, farthest from $A$. $\Gamma_1$ is the common tangent line of $\odot(AZ_2B)$ and $\odot(AZ_1B)$, farthest from $A$. Let $E = \Gamma_1 \cap \ell$, $F = \Gamma_2 \cap \ell$, and $D = \Gamma_1 \cap \Gamma_2$. Prove that $D \in \odot(ABE)$ and $\odot(ACF)$. We will only prove $D \in \odot(ACF)$, because the other case is similar. The gist is the following claim: Claim: $\odot(ABZ_1) \cup \odot(ABZ_2) \cup \Gamma_2 \sim \odot(ACZ_1) \cup \odot(ACZ_2) \cup \Gamma_1$. Proof. Let $T_1, T_2, T_3, T_4$ be the center of $\odot(AZ_1B), \odot(AZ_2B), \odot(AZ_1C), $ and $\odot(AZ_2C)$, respectively. It suffices to show that $AT_1BT_2 \sim AT_3CT_4$. It is true by the following angle chasing: $\measuredangle AT_1B = 2\measuredangle AZ_1B = 2\measuredangle AZ_1C = \measuredangle AT_1C$ and similarly $\measuredangle AT_2B = 2\measuredangle AZ_2B = 2 \measuredangle AZ_2C = \measuredangle AT_4C.$ $\blacksquare$ It is easy to see that the figures of the claim above are similar with $A$ as their center of spiral similarity (which brings $B$ to $C$). Let $I$ be the image of $F$ under this spiral similarity. Note that: \[ \measuredangle DIC = \measuredangle EIC = \measuredangle(\Gamma_1, IC) = \measuredangle(\Gamma_2, FB) = \measuredangle DFC, \]so $D, F, C, I$ are concyclic. But from the spiral similarity we also have $\measuredangle AFC = \measuredangle AFB = \measuredangle AIC$, which implies $A, F, C, I$ are concyclic. Therefore, $D \in \odot(ACF)$, as desired. $\square$.

Very cool, solved with primesarespecial Let $Y=\overline{XD} \cap \Gamma_2$. Note that $\measuredangle AXY = \measuredangle AXD = \measuredangle AFD = \measuredangle AFC = \measuredangle ABC$. $\measuredangle AYX = \measuredangle AYD = \measuredangle AED = \measuredangle AEB = \measuredangle ACB$. Hence $A$ is the center of spiral similarity mapping $XY \mapsto BC$ so it also maps $XB \mapsto YC$ and \[\measuredangle YCA = \measuredangle XBA = \measuredangle ZBA = \measuredangle ZCA\]meaning $Z, Y, C$ are collinear. To finish, note that \[\measuredangle AYC = \measuredangle AXB = \measuredangle ADX = \measuredangle ADY. \]Motivation for point YAt first it was really hard to squeeze out any information out of the problem because of the messy diagram and the fact the points look so close to eachother. Once you add the "second points" $X'$ and $Y'$ with analogous definitions and given the problem statement you know that the result should be true for them as well. Then notice the angle between tangents $XBX'$ and $YCY'$ should be equal because $Z$ lies on $(ABC)$ and from there the problem unravels itself in the form of spiral similarities. Unfortunately we had to take a sneak peek of this "equal angles" fact on geogebra because of skill issue (: