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 $△ADF\sim △GCF$. Hence

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

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

In view of ① and ②, $\frac{CL}{CK}=\frac{EL}{EK}$, this means $△CLE\sim △CKE$. 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.