silouan wrote: Given is a parallelogram $ABCD$, with $AB <AC <BC$. Points $E$ and $F$ are selected on the circumcircle $\omega$ of $ABC$ so that the tangenst to $\omega$ at these points pass through point $D$ and the segments $AD$ and $CE$ intersect. It turned out that $\angle ABF = \angle DCE$. Find the angle $\angle{ABC}$. Can this text be read as Points $E$ and $F$ are selected on the circumcircle $\omega$ of $ABC$ so that the tangents to $\omega$ at these points pass through point $D$ and through the intersection of the segments $AD$ and $CE$ ?

No, the text is as it was given. The meaning of the condition that $AD$ and $CE$ intersect is to avoid different configurations.

Denote by $O$ the center of $\omega.$ Parallel from $B$ to $AC$ cuts $DC,DA$ at $Y,Z.$ Since $\omega$ becomes 9-point circle of $\triangle DYZ,$ then it cuts $DC$ again at the projection $P$ of $Z$ on $DC.$ Since $\angle ABF=\angle DCE$ $\Longrightarrow$ $AF=EP$ $\Longrightarrow$ $AEFP$ is isosceles trapexoid with bases $EF \parallel AP$ $\Longrightarrow$ $DO \perp (EF \parallel AP)$ $\Longrightarrow$ $DO$ is perpendicular bisector of $\overline{AP}$ $\Longrightarrow$ $DP=DA=AZ=AP$ $\Longrightarrow$ $\triangle DAP$ is equilateral $\Longrightarrow$ $\angle ABC=\angle ADP=60^{\circ}.$

Another way to prove: Just an outline: 1. Extend $DC$ so that it intersect circle $\omega$ at $G$, we have $\angle GCF=\angle ACE$; so arc $AE=$arc $FG$ 2. From arc$AE=$ arc $FG$, we can deduce that $\angle ADE=\angle CDF$, so $\triangle ECD \sim \triangle AFD$ 3. Let the intersection of $AD$ and $\omega$ be $H$. From $\triangle ECD \sim \triangle AFD$, we can deduce that $DC=DH$. 4. From $DC=DH$, one can finally get $\angle ABC=\frac{\pi}{3}$.

Here is a solution without adding any additional points, though it needs trig and is actually not as simple to think of as it might seem. Denote $\angle ABC = x$, $\angle CBF = y$ and $\angle ACE = z$. Then $\angle DCE = x+y$, so $\angle ACB = 180^{\circ} - 2x - y - z$, $\angle BCF = x+z$ (from the circle of $AFCB$) and $\angle FCD = 180^{\circ} - z$ with $\angle CFD = y$. On the other hand, $\angle AED$ is adjacent to the peripheral angle corresponding to arc $AE$, so $\angle AED = 180^{\circ} - z = \angle FCD$; we also have $\angle EAD = \angle BAE - \angle BAD = 180^{\circ} - \angle BCE - \angle BAD = 3x + y - 180^{\circ}$. Therefore $x=60^{\circ}$ is equivalent to $\angle EAD = \angle CFD$ which together with $\angle FCD = \angle AED$ is equivalent to proving $\triangle FCD \sim \triangle AED$, i.e. $\frac{AE}{ED} = \frac{FC}{CD}$. From the SIne Law in triangle $FCD$ and the tangents $DE = DF$ we have $\frac{DE}{CD} = \frac{DF}{CD} = \frac{\sin z}{\sin y}$, and from the Sine Law in the circumcircle of $ABC$ we have $\frac{AE}{FC} = \frac{\sin \angle ACE}{\sin \angle CBF} = \frac{\sin z}{\sin y}$. This completes the proof.