Bulgarian Spring Tournament

Let the tangent to $k$ at point $C$ intersect the rays $AB \to$ and $AD \to$ at the points $M$ and $N$, respectively, and the lines $BD$, $EF$ and the tangent intersect at point $P$. After applying Menelaus' theorem twice to $\triangle AMN$ and to the lines $BD$ and $EF$, we get
$$\frac{AD}{ND} \cdot \frac{NP}{MP} \cdot \frac{MB}{AB} = 1 \quad \text{and} \quad \frac{AF}{NF} \cdot \frac{NP}{MP} \cdot \frac{ME}{AE} = 1.$$
Hence
$$(1) \quad \frac{AD}{ND} \cdot \frac{MB}{AB} = \frac{AF}{NF} \cdot \frac{ME}{AE}.$$
Since $ABCD$ is a parallelogram, then $\frac{AD}{ND} = \frac{MC}{NC} = \frac{MB}{AB}$ (2). From the tangent and secant property $MC^2 = ME \cdot MA$ and $NC^2 = NF \cdot NA$ (3). From (1), (2) and (3) it follows that
$$\frac{MC^2}{NC^2} = \frac{AF}{NF} \cdot \frac{ME}{AE} \Rightarrow \frac{ME \cdot MA}{NF \cdot NA} = \frac{AF}{NF} \cdot \frac{ME}{AE}.$$
Therefore, $AM \cdot AE = AN \cdot AF$, i.e. the quadrilateral $EFNM$ is cyclic. Then $\angle AEF = \angle ANM$, whence $\overline{AF} = \overline{AEC} - \overline{FC}$, i.e., $\overline{AEC} = \overline{AFC}$. It follows that $AC$ is a diameter of $k$.