Assume that a part of ∆ABC around vertex A is not visible. Describe how to find the angle bisector of ∠CAB. Assume that a part of ∆ABC around vertex A is not visible. Describe how to find
the angle bisector of ∠CAB.
I have no idea how to begin this problem. Any help is appreciated. Thank you. 

You may exploit the following property of the angle bisector through $A$: for any point $P$ on it, the distance of $P$ from the $AB$-side equals the distance of $P$ from the $AC$-side. Just find two distinct points with such property in the visible part and join them to get the wanted line.

You may take $P=I$ as the intersection of the internal angle bisectors from $B$ and $C$ and $Q=I_A$ as the intersection of the external angle bisectors from $B$ and $C$, for instance.
An alternative approach through triangle similarities:



*

*Take $B'\in AB$ close to $B$ and let $C'\in AC$ such that $B'C'\parallel BC$;

*Let $C''\in BC$ be such that $B' C''\parallel AC$;

*Let $J$ the point in which the angle bisector of $\widehat{BB'C''}$ meets the $BC$-side and $K$ the projection of $J$ on $B'C''$;

*Let $L=BK\cap AC$ and $N\in BC$ be such that $LN\parallel KJ$;

*The parallel to $B'J$ through $L$ is the wanted angle bisector.


A third approach:



*

*Let $B'$ and $C'$ like before, let $M$ be the midpoint of $B'C'$;

*Let $A'$ be the intersection between the parallel to $AC$ through $B'$ and the parallel to $AB$ through $C'$;

*Let $J\in B'C'$ be on the angle bisector of $\widehat{B' A' C'}$;

*Let $K$ be the symmetric of $J$ with respect to $M$;

*The wanted line is the parallel to $A'J$ through $K$.

