Subfield: Geometry


Question: Let $A B C D$ be a cyclic quadrilateral whose sides have pairwise different lengths. Let $O$ be the circumcentre of $A B C D$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $B_{1}$ and $D_{1}$, respectively. Let $O_{B}$ be the centre of the circle which passes through $B$ and is tangent to $A C$ at $D_{1}$. Similarly, let $O_{D}$ be the centre of the circle which passes through $D$ and is tangent to $A C$ at $B_{1}$.

Assume that $B D_{1} \| D B_{1}$. Prove that $O$ lies on the line $O_{B} O_{D}$.

Solution:
We introduce some objects and establish some preliminary facts common for all solutions below.

Let $\Omega$ denote the circle $(A B C D)$, and let $\gamma_{B}$ and $\gamma_{D}$ denote the two circles from the problem statement (their centres are $O_{B}$ and $O_{D}$, respectively). Clearly, all three centres $O, O_{B}$, and $O_{D}$ are distinct.

Assume, without loss of generality, that $A B>B C$. Suppose that $A D>D C$, and let $H=A C \cap B D$. Then the rays $B B_{1}$ and $D D_{1}$ lie on one side of $B D$, as they contain the midpoints of the arcs $A D C$ and $A B C$, respectively. However, if $B D_{1} \| D B_{1}$, then $B_{1}$ and $D_{1}$ should be separated by $H$. This contradiction shows that $A D<C D$.

Let $\gamma_{B}$ and $\gamma_{D}$ meet $\Omega$ again at $T_{B}$ and $T_{D}$, respectively. The common chord $B T_{B}$ of $\Omega$ and $\gamma_{B}$ is perpendicular to their line of

<img_3842>
centres $O_{B} O$; likewise, $D T_{D} \perp O_{D} O$. Therefore, $O \in O_{B} O_{D} \Longleftrightarrow O_{B} O\left\|O_{D} O \Longleftrightarrow B T_{B}\right\|$ $D T_{D}$, and the problem reduces to showing that

$$
B T_{B} \| D T_{D}
\tag{1}
$$


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

$$
\frac{H L_{B}}{\sin \angle H B L_{B}}=\frac{B L_{B}}{\sin \angle B H L_{B}}=\frac{B_{1} L_{B}}{\sin \angle B H L_{B}}
\tag{2}
$$

similarly,

$$
\frac{H L_{D}}{\sin \angle H D L_{D}}=\frac{D L_{D}}{\sin \angle D H L_{D}}=\frac{D_{1} L_{D}}{\sin \angle D H L_{D}}
\tag{3}
$$

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.



<img_3647>
Let $B D_{1}$ and $T_{B} D_{1}$ meet $\Omega$ again at $X_{B}$ and $Y_{B}$, respectively. Then

$$
\angle B D_{1} C=\angle B T_{B} D_{1}=\angle B T_{B} Y_{B}=\angle B X_{B} Y_{B}
$$

which shows that $X_{B} Y_{B} \| A C$. Similarly, let $D B_{1}$ and $T_{D} B_{1}$ meet $\Omega$ again at $X_{D}$ and $Y_{D}$, respectively; then $X_{D} Y_{D} \| A C$.

Let $M_{D}$ and $M_{B}$ be the midpoints of the arcs $A B C$ and $A D C$, respectively; then the points $D_{1}$ and $B_{1}$ lie on $D M_{D}$ and $B M_{B}$, respectively. Let $K$ be the midpoint of $A C$ (which lies on $M_{B} M_{D}$ ). Applying Pascal's theorem to $M_{D} D X_{D} X_{B} B M_{B}$, we obtain that the points $D_{1}=M_{D} D \cap X_{B} B, B_{1}=D X_{D} \cap B M_{B}$, and $X_{D} X_{B} \cap M_{B} M_{D}$ are collinear, which means that $X_{B} X_{D}$ passes through $K$. Due to symmetry, the diagonals of an isosceles trapezoid $X_{B} Y_{B} X_{D} Y_{D}$ cross at $K$.

<img_3168>

Let $b$ and $d$ denote the distances from the lines $X_{B} Y_{B}$ and $X_{D} Y_{D}$, respectively, to $A C$. Then we get

$$
\frac{X_{B} Y_{B}}{X_{D} Y_{D}}=\frac{b}{d}=\frac{D_{1} X_{B}}{B_{1} X_{D}}
$$

where the second equation holds in view of $D_{1} X_{B} \| B_{1} X_{D}$. Therefore, the triangles $D_{1} X_{B} Y_{B}$ and $B_{1} X_{D} Y_{D}$ are similar. The triangles $D_{1} T_{B} B$ and $B_{1} T_{D} D$ are similar to them and hence to each other. Since $B D_{1} \| D B_{1}$, these triangles are also homothetical. This yields $B T_{B} \| D T_{D}$, as desired.