, 2010

Let the lines $DE$ and $CA$ meet at point $L$. We will prove that the line $BK$ passes through $L$ (see figure 1).
The circle $c(O, R)$ is homothetic to the circle $c_1(K, R_1)$ with respect to homothety with center $A$ and ratio $m = \frac{R}{R_1}$, say $H(A, \frac{R}{R_1})$. The extension of $CD$ meets the circle ($c$) at a point $D_1$ homothetic of $D$. The extension of $CE$ meets the circle ($c$) at a point $E_1$ homothetic of $E$.
Therefore the line segment $CE_1$ is homothetic of the line segment $CE$. So, if the line $AC$ intersects $D_1E_1$ at the point $L_1$, then $L_1$ will be homothetic of $L$. Since $O$ is homothetic of $K$, we conclude that
$$OL_1 \parallel KL. \quad (1)$$
We will prove that
$$OL_1 \parallel BL. \quad (1)$$
Since $AD$ and $AE$ are tangents from $A$ to the circle ($c_1$), then $AK$ is the perpendicular bisector of the segment $DE$. Let $M$ be the intersection point of the lines $AK$ and $DE$, such that the extension of $CM$ intersects $D_1E_1$ at $M_1$ and the circle ($c$) at point $M_2$. Then $M_1$ will be the middle point of the segment $D_1E_1$ (because of the homothety).
We assert that $CA$ is the symmedian of the triangle $CDE$ which corresponds to the vertex $C$. According to Steiner's theorem to symmedians, it is enough to prove that
$$\frac{DL}{LE} = \frac{CD^2}{CE^2}. \quad (3)$$
For proving the relation (3) we use the areas ratio:
$$\frac{\sigma(CDL)}{\sigma(CEL)} = \frac{DL}{LE} = \frac{\sigma(DAL)}{\sigma(EAL)} = \frac{\sigma(CDL) + \sigma(DAL)}{\sigma(CEL) + \sigma(EAL)} = \frac{\sigma(CAD)}{\sigma(CAE)}. \quad (4)$$
Since the angles $ADE$ and $AED$ are the angles between tangents and chord we have
$\angle ADE = \angle AED = \angle DCE$ and therefore
$$\angle CDA = \angle CDE + \angle ADE = \angle CDE + \angle DCE = 180^\circ - \angle CED,$$
$$\angle CEA = \angle CED + \angle AED = \angle CED + \angle DCE = 180^\circ - \angle CDE.$$
From (4) we obtain
$$\frac{DL}{LE} = \frac{\sigma(CAD)}{\sigma(CAE)} = \frac{CD \cdot \sin(180^\circ - \angle CED)}{CE \cdot \sin(180^\circ - \angle CDE)} = \frac{CD \cdot \sin(\angle CED)}{CE \cdot \sin(\angle CDE)} = \frac{CD^2}{CE^2}.$$

Figure 1
So, the relation (3) is proved and $CA$ is the symmedian of the triangle $CDE$ which corresponds to the vertex $C$.
Hence $\angle D_1CA = \angle E_1CM_2$ and the quadrilateral $AD_1E_1M_2$ is an isosceles trapezium. The line $OM_1$ is perpendicular to $D_1E_1$ and intersects $AM_2$ at the middle $N$. In the triangle $AMM_2$ we have that $N$ is the middle of the side $AM_2$ and $NM_1 \parallel AM$. Hence $M_1$ is the middle of $MM_2$ and therefore $D_1E_1$ is the mid-parallel of $AM_2$ and $DE$. Since $L_1$ belongs to $D_1E_1$, it will be the middle of $AL$. In the triangle $ALB$ $OL_1$ is the mid-parallel to $BL$. Hence $OL_1 \parallel BL$ and the straight lines $AC$, $BK$ and $DE$ are concurrent. $\square$