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 $△AMN$ and to the lines $BD$ and $EF$, we get $$\frac{AD}{ND}\cdot \frac{NP}{MP}\cdot \frac{MB}{AB}=1\text{and}\frac{AF}{NF}\cdot \frac{NP}{MP}\cdot \frac{ME}{AE}=1.$$ Hence $$(1)\frac{AD}{ND}\cdot \frac{MB}{AB}=\frac{AF}{NF}\cdot \frac{ME}{AE}.$$ Since $ABCD$ is a parallelogram, then $\frac{AD}{ND}=\frac{MC}{NC}=\frac{MB}{AB}$ (2). From the tangent and secant property $M{C}^2=ME\cdot MA$ and $N{C}^2=NF\cdot NA$ (3). From (1), (2) and (3) it follows that $$\frac{M{C}^2}{N{C}^2}=\frac{AF}{NF}\cdot \frac{ME}{AE}\Rightarrow \frac{ME\cdot MA}{NF\cdot NA}=\frac{AF}{NF}\cdot \frac{ME}{AE}.$$ 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 $k$.