Do you mean tangent by touches the circle $\omega$ cause I think it cannot be a tangent(only in a specific case).

I have some proof(that it intersects) which I hope is correct. I am giving a partial solution right now and will complete when I have time. Obviously $AIC$ is a straight line. Also it is pretty well known $CD=CI=CB$. Let the circle passing through $D,I,B$ with $C$ centre be $\omega_1$ Now consider the homothety $H_C:I \to A$ that is centred at $C$ which takes $I$ to $A$. We will show $H_C:\omega \to \omega_1$.

consider$I_e$ the excenter ,T the projection of C on the parallel to AB through I X= $\frac{R-r}{2},tan \frac{A}{2}=\frac{r}{s-a}=\frac{R}{s}=\frac{r'}{\frac{a}{2}}$,where[a=BD,s=$ \frac{BD+AB+AD}{2}$] the consequence is obvious

Attachments:

Firstly, $ A,I,C $ are colinear and $ BC=IC=DC $ And let $ R $ be the tangent point of $ \omega $ with $ BD $ . Clearly, $ BR=RD $ and now let $ Q $ be the pedal of projection from $ C $ to the straight line through $ I $ and parallel to $ AB $ . Now, consider, $ \triangle IQC $ and $ \triangle BRC $ : We have: $ BC=IC $ , $ \angle CBR=\angle QIC=\angle A/2 $ . And now as they are right angled so, $ \angle BCR=\angle ICQ $ .so they are congruent and finally, $ CQ=CR $

I solved this problem,but i don't know Latex inthedarkness wrote: Firstly, $ A,I,C $ are colinear and $ BC=IC=DC $ And let $ R $ be the tangent point of $ \omega $ with $ BD $ . Clearly, $ BR=RD $ and now let $ Q $ be the pedal of projection from $ C $ to the straight line through $ I $ and parallel to $ AB $ . Now, consider, $ \triangle IQC $ and $ \triangle BRC $ : We have: $ BC=IC $ , $ \angle CBR=\angle QIC=\angle A/2 $ . And now as they are right angled so, $ \angle BCR=\angle ICQ $ .so they are congruent and finally, $ CQ=CR $

So your method is same as me? Congrats!

So your method is same as me? Congrats!

Let $IX$ be the tangent from $I$ to the circle $\omega$ and $X$ lies on $\omega$ and let $\omega$ touches $BD$ at $M$. The equality $CB = CI = CD$ is well-known and easily proven, then the triangles $\triangle{CXI} \cong ruent \triangle{CMB}$. So, $\angle{CAB} = \angle{CBM} = \angle{CIX}$. So, $IX$ is parallel to $AB$. Thus, $IX$ is the line through $I$ and parallel to $AB$.