First, notice that $ K$ and $ L$ are both either on or both outside the segment $ AB$. Thus, it is sufficient to prove that $ AK = LB$ and to prove this, it is obviously sufficient to prove $ \angle ARK = \angle LEB$, where $ R$ is the tangency point on $ AC$ to the incircle of $ \triangle ABC$. Let $ I$ be the incenter of $ \triangle ABC$. Then, \[ \angle EFR = \frac {\angle EIR }2 = \angle CIR = \angle BAC, \] so $ A,K,F,R$ are cyclic. Thus, \[ \angle ARK = \angle AFK = \angle GFE = \angle LEB, \] which solves the problem.

It is also possible to use Menelaus and power of a point, based on a solution of czech contestant Tomas Pavlik.

Yes, I know, but I can't help my self. Although there is solution already posted I will post my solution. It is nothing special but it is projective. Let incircle touch $ AC$ at $ T$ and let $ TD$ cut $ FG$ at $ X$ Since $ TD$ is polar line for $ A$ we have $ (F,G;X,A)=-1$. Now we have: \begin{eqnarray*} (K,L;D,\infty) &=& (EK,EL;ED,E\infty)\\ &=& (EF,EG;ED,ET)\\ &=& (F,G;D,T)\\ &=& (TF,TG;TD,TT)\\ &=& (TF,TG;TX,TA)\\ &=& (F,G;X,A)=-1 \end{eqnarray*} So $ D$ is midpoint of segment $ KL$. Q.E.D.

Dear Number1, we have quadrilateral $ TFDG$ is harmonic, so $ E.(TFDG)$ is also a harmonic bundle. Now, notice that $ ET\parallel KL$, the result follows.

Every time there are two equal segments next to each other, there is some kind of asynchronous ( possibly) degenerate species of butterfly flying around. M.T.

The solution using Menelaus theorem according to Tomas Pavlik: Menelaus theorem is used twice to the triangle $AXB$ and lines $EF$ and $GE$, where $X$ is the intersection of $AF$ and $BC$. You will get to $AL*AK=BK*BL$ which we can somehow finish.

Let $T$ be the tangency point of $AC$ with the incircle. Since we have that $BC=CA$ this implies $ET \parallel BC$, thus we have that: $$-1=(F,G;D,T) \overset{E}{=} (K,L;D,P_{\infty})$$this gives us $KD=DL$ ...