Transitivity of containment of angles. Suppose the ray AD is in the interior of
the $\angle BAC$, and the ray AE is in
the interior of the $ \angle DAC$. Show
that AE is also in the interior of $ \angle BAC$. Using Hilbert's axioms this is exercise 9.2 in Hartshorne's book Geometry: Euclid and Beyond pg 96.
I have alot of trouble with proving rigorously really obvious things in mathematics i imagine this exercise isn't meant to be that difficult but i cant seem to figure out where to start.


Notions from chapter 7 "Axioms of betweenness" of the book would be helpful here.
From the same pdf I linked in one of your earlier previous question, you have the definition:

The inside (or interior) of an angle $\angle BAC$ consists of all points
  $D$ such that $D$ and $C$ are on the same side of the line $AB$, and $B$ and $B$ are on the same side of the line $AC$.

Note that earlier in the chapter they properly define what it means to be "on the same side" of a line.
Using Proposition 7.3 (Crossbar theorem) that comes right after the passage I quoted, you should be able to prove that:

A ray $AD$ is inside the angle $\angle BAC$ if and only if ray $AD$ meets segment $BC$.

The crossbar theorem gives you one implication.
Conversely suppose 
$\angle BAC$ is an angle and there is some point $D'$ on ray $AD$ such that
$B\ast D'\ast C$.
Then $D'$ and $C$ are on the same side of line $AB$.
Because line $AB$ intersects line $AD$ at point $A$, it splits line $AD$ into two rays, one on either side of line $AB$. One of these rays is precisely ray $AD$. That ray $AD$ contains $D'$ by assumption, so $D$ and $D'$ are on the same side of line $AB$, it follows that $D$ and $C$ are on the same side of line $AB$. Similarly, $D$ and $B$ are on the same side of line $AC$. Therefore, ray $AD$ is inside angle $\angle BAC$.
I think the above proof holds with your axioms. I didn't read chapter 7 in detail but it contains stuff about "plane separation" (by a line) and "line separation" (by a point), so I assume most of the intuition about "betweenness" should be available when they define the interior of an angle.

From there a possible proof for your exercise would be to use the crossbar theorem twice:


*

*find $D'$ on ray $AD$ and segment $BC$ (the point you called $H$ in the comments) 

*then you can get a point $E'$ on ray $AE$ and segment $D'C$ (the point you called $F$)


Then you have $B\ast D'\ast C$ and $D'\ast E'\ast C$, from which you should be able to deduce $B\ast E'\ast C$ from some result in the book. At least I hope something like that is in there.
You can conclude with the characterization of a ray inside an angle that I gave above.

Is there something simpler/faster? Maybe, I haven't checked everything in the book. Maybe there's a result related to my characterization of a ray inside an angle somewhere in the book that you can re-use to go faster. One suspicious place would be the exercises in chapter 7 (that I didn't check out).
