Problem 1614

Let the diagonals $A C$ and $B D$ cross at $H$. Consider the homothety $h$ centred at $H$ and mapping $B$ to $D$. Since $B D_{1} \| D B_{1}$, we have $h\left(D_{1}\right)=B_{1}$. Let the tangents to $\Omega$ at $B$ and $D$ meet $A C$ at $L_{B}$ and $L_{D}$, respectively. We have
$$\angle L_{B} B B_{1}=\angle L_{B} B C+\angle C B B_{1}=\angle B A L_{B}+\angle B_{1} B A=\angle B B_{1} L_{B},$$
which means that the triangle $L_{B} B B_{1}$ is isosceles, $L_{B} B=L_{B} B_{1}$. The powers of $L_{B}$ with respect to $\Omega$ and $\gamma_{D}$ are $L_{B} B^{2}$ and $L_{B} B_{1}^{2}$, respectively; so they are equal, whence $L_{B}$ lies on the radical axis $T_{D} D$ of those two circles. Similarly, $L_{D}$ lies on the radical axis $T_{B} B$ of $\Omega$ and $\gamma_{B}$. By the sine rule in the triangle $B H L_{B}$, we obtain
$$\frac{H L_{B}}{\sin \angle H B L_{B}}=\frac{B L_{B}}{\sin \angle B H L_{B}}=\frac{B_{1} L_{B}}{\sin \angle B H L_{B}}$$
similarly,
$$\frac{H L_{D}}{\sin \angle H D L_{D}}=\frac{D L_{D}}{\sin \angle D H L_{D}}=\frac{D_{1} L_{D}}{\sin \angle D H L_{D}}$$
Clearly, $\angle B H L_{B}=\angle D H L_{D}$. In the circle $\Omega$, tangent lines $B L_{B}$ and $D L_{D}$ form equal angles with the chord $B D$, so $\sin \angle H B L_{B}=\sin \angle H D L_{D}$ (this equality does not depend on the picture). Thus, dividing (2) by (3) we get
$$\frac{H L_{B}}{H L_{D}}=\frac{B_{1} L_{B}}{D_{1} L_{D}}, \quad \text { and hence } \quad \frac{H L_{B}}{H L_{D}}=\frac{H L_{B}-B_{1} L_{B}}{H L_{D}-D_{1} L_{D}}=\frac{H B_{1}}{H D_{1}}$$
Since $h\left(D_{1}\right)=B_{1}$, the obtained relation yields $h\left(L_{D}\right)=L_{B}$, so $h$ maps the line $L_{D} B$ to $L_{B} D$, and these lines are parallel, as desired. !

Comment 2. In the solution above, the key relation $h\left(L_{D}\right)=L_{B}$ was obtained via a short computation in sines. Here we present an alternative, pure synthetical way of establishing that. Let the external bisectors of $\angle A B C$ and $\angle A D C$ cross $A C$ at $B_{2}$ and $D_{2}$, respectively; assume that $\overparen{A B}>\overparen{C B}$. In the right-angled triangle $B B_{1} B_{2}$, the point $L_{B}$ is a point on the hypothenuse such that $L_{B} B_{1}=L_{B} B$, so $L_{B}$ is the midpoint of $B_{1} B_{2}$. Since $D D_{1}$ is the internal angle bisector of $\angle A D C$, we have
$$\angle B D D_{1}=\frac{\angle B D A-\angle C D B}{2}=\frac{\angle B C A-\angle C A B}{2}=\angle B B_{2} D_{1},$$
so the points $B, B_{2}, D$, and $D_{1}$ lie on some circle $\omega_{B}$. Similarly, $L_{D}$ is the midpoint of $D_{1} D_{2}$, and the points $D, D_{2}, B$, and $B_{1}$ lie on some circle $\omega_{D}$. Now we have
$$\angle B_{2} D B_{1}=\angle B_{2} D B-\angle B_{1} D B=\angle B_{2} D_{1} B-\angle B_{1} D_{2} B=\angle D_{2} B D_{1} .$$
Therefore, the corresponding sides of the triangles $D B_{1} B_{2}$ and $B D_{1} D_{2}$ are parallel, and the triangles are homothetical (in $H$). So their corresponding medians $D L_{B}$ and $B L_{D}$ are also parallel. !

Yet alternatively, after obtaining the circles $\omega_{B}$ and $\omega_{D}$, one may notice that $H$ lies on their radical axis $B D$, whence $H B_{1} \cdot H D_{2}=H D_{1} \cdot H B_{2}$, or
$$\frac{H B_{1}}{H D_{1}}=\frac{H B_{2}}{H D_{1}} .$$
Since $h\left(D_{1}\right)=B_{1}$, this yields $h\left(D_{2}\right)=B_{2}$ and hence $h\left(L_{D}\right)=L_{B}$.

Comment 3. Since $h$ preserves the line $A C$ and maps $B \mapsto D$ and $D_{1} \mapsto B_{1}$, we have $h\left(\gamma_{B}\right)=\gamma_{D}$. Therefore, $h\left(O_{B}\right)=O_{D}$; in particular, $H$ also lies on $O_{B} O_{D}$.