According to this property, $$\frac{\sin\angle ADK}{\sin\angle KDB}\times\frac{AD}{DB}=\frac{AK}{KB},$$In addition, $$\frac{\sin\angle ADL}{\sin\angle LDC}\times\frac{AD}{CD}=\frac{AL}{CL}.$$Notic that $\frac{AL}{LC}=\frac{AK}{KB}$. Combining them we get that $$\frac{\sin\angle ADK}{\sin\angle KDB}\times\frac{AD}{DB}=\frac{\sin\angle ADL}{\sin\angle LDC}\times\frac{AD}{CD}.$$Let $\angle KDB=\alpha$, $\angle LDC=\beta$. Then $$\frac{\sin(\angle ADB-\alpha)}{\sin\alpha}\times\frac{CD}{DB}=\frac{\sin(\angle ADC-\beta)}{\sin\beta}.$$Applying the law of sines, $\frac{CD}{DB}=\frac{\sin\angle CBD}{\sin\angle BCD}$. Here $$\angle ADB-\alpha=\angle KDA=\angle BCD$$and \begin{align*}&\pi-\angle CBD=\angle BCD+\angle CDB\\=&\angle ADK+\angle CDB=\angle ADC-\beta.\end{align*}Notice that $\sin(\pi-\angle CBD)=\sin\angle CBD=\sin(\angle ADC-\beta)$, and plug in. Canceling out terms will leave us with $$\sin\alpha=\sin\beta.$$The fact that $\alpha+\beta=\pi$ cannot be true because $$\angle KDB=\angle LDC<\angle ADC<\pi.$$So $\alpha=\angle KDB=\angle LDC=\beta$.

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