International Mathematical Olympiad

Draw the altitudes of two isosceles triangles $EGC$ and $ECF$ as in the figure.
In view of the given condition, it is easy to see that $\triangle ADF \sim \triangle GCF$. Hence

$$\begin{aligned} \frac{AD}{GC} &= \frac{DF}{CF} \Rightarrow \frac{BC}{CG} = \frac{DF}{CF} \Rightarrow \frac{BC}{CL} = \frac{DF}{CK} \\ &\Rightarrow \frac{BC + CL}{CL} = \frac{DF + FK}{CK} \\ &\Rightarrow \frac{BL}{CL} = \frac{DK}{CK} \\ &\Rightarrow \frac{BL}{DK} = \frac{CL}{CK}. \end{aligned} \quad ①$$

Since $BCED$ is a cyclic quadrilateral, $\angle LBE = \angle EDK$, this yields $\triangle BLE \sim \triangle DKE$, where both are right-angled triangles.
$$\text{So} \qquad \frac{BL}{DK} = \frac{EL}{EK}. \qquad ②$$

In view of ① and ②, $\frac{CL}{CK} = \frac{EL}{EK}$, this means $\triangle CLE \sim \triangle CKE$.
Thus
$$\frac{CL}{CK} = \frac{CE}{CE} = 1,$$
i.e. $CL = CK \Rightarrow CG = CF$.
It is intuitively obvious that $\angle BAG = \angle GAD$. Hence $l$ is the bisector.