Estonia — Open Contests

Fig. 9

Fig. 10

Let $B'$ and $C'$ be respectively the second intersection points of the lines $AB$ and $AC$ with the circumcircle of the triangle $BCK$ (Fig. 9). Notice that $\angle BKC = 180^\circ - \angle CBK - \angle BCK = 180^\circ - \frac{\angle ABC}{2} - (90^\circ + \frac{\angle ACB}{2}) = 90^\circ - \frac{\angle ABC}{2} - \frac{\angle ACB}{2}$. Hence the central angle supported by the arc $BC$ has the size $180^\circ - \angle ABC - \angle ACB$. The central angle supported by the arc $B'C$ has the size $2\angle ABC$ and the central angle supported by the arc $C'B$ has the size $2\angle ACB$. Hence the central angle supported by the arc $B'C'$ has the size $360^\circ - (180^\circ - \angle ABC - \angle ACB) - 2\angle ABC - 2\angle ACB$, which equals $180^\circ - \angle ABC - \angle ACB$. Hence the arcs $BC$ and $B'C'$ have the same size. As $K$ is the midpoint of the arc $B'C$, the point $B'$ is the reflection of the point $C$ by the diameter drawn from point $K$, and $C'$ is the reflection of point $B$. Hence the intersection point $A$ of $BB'$ and $CC'$ has to lie on the diameter drawn from the point $K$.

Remark. Among the used central angles there may also be angles of size greater than $180^\circ$ (reflex angles), such as the angle corresponding to the arc $B'C$ in Fig. 10.

Fig. 11

Let $L$ be the second endpoint of the diameter through $K$ of the circumcircle of the triangle $BCK$ (Fig. 11). Then $\angle KCL = 90^\circ$, thus $CL$ is the bisector of the interior angle at vertex $C$ of the triangle $ABC$, and $\angle KBL = 90^\circ$, so $BL$ is the bisector of the exterior angle at vertex $B$ of $ABC$.
The bisectors of the exterior angles at some two vertices of a triangle and the bisector of the interior angle at the third vertex of the triangle intersect at a common point. Thus the bisector of the exterior angle at vertex $A$ of the triangle $ABC$ passes through points $K$ and $L$. Thus point $A$ lies on $KL$.

Fig. 12

Let us denote the angles of the triangle $ABC$ at vertices $A$, $B$ and $C$ respectively $\alpha$, $\beta$ and $\gamma$. Let the intersection point of the line $BC$ and the tangent to the circumcircle of triangle $BCK$ at point $K$ be $M$ (Fig. 12). By the property of inscribed angles, $\angle MKC = \angle MBK = \frac{\beta}{2}$. The bisectors of the exterior angles at some two vertices of a triangle and the bisector of the interior angle at the third vertex of the triangle have a common point. Hence $AK$ is the bisector of the exterior angle at vertex $A$ of the triangle $ABC$.
---
Thus $\angle CKA = 180^\circ - \angle KAC - \angle KCA = 180^\circ - \frac{180^\circ - \alpha}{2} - \frac{180^\circ - \gamma}{2} = \frac{\alpha}{2} + \frac{\gamma}{2}$.
Hence $\angle MKA = \angle MKC + \angle CKA = \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^\circ$. As a tangent of a circle is perpendicular to the diameter drawn from the point of tangency we have that $A$ lies on the diameter drawn from point $K$.