, 2011

Since $\omega$ and the circumcircle of triangle $A C E$ are concentric, the tangents from $A$, $C$, and $E$ to $\omega$ have equal lengths; that means that $A B=B C, C D=D E$, and $E F=F A$. Moreover, we have $\angle B C D=\angle D E F=\angle F A B$.

Consider the rotation around point $D$ mapping $C$ to $E$; let $B^{\prime}$ and $L^{\prime}$ be the images of the points $B$ and $J$, respectively, under this rotation. Then one has $D J=D L^{\prime}$ and $B^{\prime} L^{\prime} \perp D E$; moreover, the triangles $B^{\prime} E D$ and $B C D$ are congruent. Since $\angle D E O<90^{\circ}$, the lines $E O$ and $B^{\prime} L^{\prime}$ intersect at some point $K^{\prime}$. We intend to prove that $K^{\prime} B \perp D F$; this would imply $K=K^{\prime}$, therefore $L=L^{\prime}$, which proves the problem statement.

Analogously, consider the rotation around $F$ mapping $A$ to $E$; let $B^{\prime \prime}$ be the image of $B$ under this rotation. Then the triangles $F A B$ and $F E B^{\prime \prime}$ are congruent. We have $E B^{\prime \prime}=A B=B C= E B^{\prime}$ and $\angle F E B^{\prime \prime}=\angle F A B=\angle B C D=\angle D E B^{\prime}$, so the points $B^{\prime}$ and $B^{\prime \prime}$ are symmetrical with respect to the angle bisector $E O$ of $\angle D E F$. So, from $K^{\prime} B^{\prime} \perp D E$ we get $K^{\prime} B^{\prime \prime} \perp E F$.

From these two relations we obtain
$$K^{\prime} D^{2}-K^{\prime} E^{2}=B^{\prime} D^{2}-B^{\prime} E^{2} \quad \text { and } \quad K^{\prime} E^{2}-K^{\prime} F^{2}=B^{\prime \prime} E^{2}-B^{\prime \prime} F^{2} .$$
Adding these equalities and taking into account that $B^{\prime} E=B^{\prime \prime} E$ we obtain
$$K^{\prime} D^{2}-K^{\prime} F^{2}=B^{\prime} D^{2}-B^{\prime \prime} F^{2}=B D^{2}-B F^{2}$$
which means exactly that $K^{\prime} B \perp D F$.

Let us denote the points of tangency of $A B, B C, C D, D E, E F$, and $F A$ to $\omega$ by $R, S, T, U, V$, and $W$, respectively. As in the previous solution, we mention that $A R= A W=C S=C T=E U=E V$.

The reflection in the line $B O$ maps $R$ to $S$, therefore $A$ to $C$ and thus also $W$ to $T$. Hence, both lines $R S$ and $W T$ are perpendicular to $O B$, therefore they are parallel. On the other hand, the lines $U V$ and $W T$ are not parallel, since otherwise the hexagon $A B C D E F$ is symmetric with respect to the line $B O$ and the lines defining the point $K$ coincide, which contradicts the conditions of the problem. Therefore we can consider the intersection point $Z$ of $U V$ and $W T$.

Next, we recall a well-known fact that the points $D, F, Z$ are collinear. Actually, $D$ is the pole of the line $U T, F$ is the pole of $V W$, and $Z=T W \cap U V$; so all these points belong to the polar line of $T U \cap V W$.

Now, we put $O$ into the origin, and identify each point (say $X$ ) with the vector $\overrightarrow{O X}$. So, from now on all the products of points refer to the scalar products of the corresponding vectors.

Since $O K \perp U Z$ and $O B \perp T Z$, we have $K \cdot(Z-U)=0=B \cdot(Z-T)$. Next, the condition $B K \perp D Z$ can be written as $K \cdot(D-Z)=B \cdot(D-Z)$. Adding these two equalities we get
$$K \cdot(D-U)=B \cdot(D-T)$$
By symmetry, we have $D \cdot(D-U)=D \cdot(D-T)$. Subtracting this from the previous equation, we obtain $(K-D) \cdot(D-U)=(B-D) \cdot(D-T)$ and rewrite it in vector form as
$$\overrightarrow{D K} \cdot \overrightarrow{U D}=\overrightarrow{D B} \cdot \overrightarrow{T D} .$$
Finally, projecting the vectors $\overrightarrow{D K}$ and $\overrightarrow{D B}$ onto the lines $U D$ and $T D$ respectively, we can rewrite this equality in terms of segment lengths as $D L \cdot U D=D J \cdot T D$, thus $D L=D J$.