I found that the relation is 3:1 but I used analitical geometry

Let the circle touches $BC$ at $E$. We have $BE=BA$ and $CE^2=CD\cdot CA$. So $CE=|a-c|$ and $(a-c)^2=\frac{a+b-c}2\cdot b$. Let $a=y+z,b=z+x,c=x+y$, we get $(x-z)^2=z(z+x)$, or $x^2-2xz+z^2=z^2+xz$, so $z^2=3zx$ and $z=3x$. So $\frac{AD}{DC}=\frac{-a+b+c}{a+b-c}=\frac{x}z=\frac13$.

Let $X=(I)\cap \overline{BD}$. Let the tangent from $X$ to $(I)$ intersect $\overline{BC}$ at $E$. Note that $(E,D;C,A)=-1$. Let $(I)$ touch $\overline{AB}$ at $R$. By homothety, $\overline{BD}$ is tangent to $(IRAD)$. Thus, by trivial angle chase, $XD=DR$. Due to congruency of $\triangle EXD$ and $\triangle DEA$, we obtain that $DA=DE$. As $(E,D;C,A)=-1$, we get that $\frac{CE}{CD}=\frac{AE}{AD}\implies CE=2CD$. Hence, $AD=DE=CD+CE=3CD\implies \frac{AD}{CD}=\boxed{3}$.

Dear, I don't understant the satments ''the latter at point A''? A picture? Sincerely Jean-Louis

Any answer? Sincerely Jean-Louis

I think it means $A$ is further from $B$ than $C$. cmiiw though