$k,l$ circles such that $k,l$ tangent to $BC$ at $B,C$ passes $A$. let other intersection of $k,l$ is $P$. let $AP \cap BC=Q$. $\angle{PBC}=\angle{BAP}=\angle{DAC}=\angle{DBC}$ similarly $\angle{PCB}=\angle{DCB}$. $P,D$ symetric to $BC$. By similitary $\frac{BD}{AB}= \frac{BP}{AB}= (\frac{QP}{QA})^\frac{1}{2}=\frac{CP}{CA}=\frac{DP}{CA}$ so $BD.AC=DC.BA$. Also we get $BD.AC+DC.BA=BC.AD$ so $2BD.AC=2DC.BA=BC.AD$. So we get $\frac{AD}{BD}+ \frac{AD}{CD}= 2\frac{ CA + AB}{CB}=6$.