Even with both statements, $\displaystyle AE$ cannot be determined because the length of $\displaystyle BD$ is missing.

For example, we can have $\displaystyle AB = DE = 8$ and $\displaystyle BD=16$, making $\displaystyle AE = 32$; or, we can have  $\displaystyle AB = DE = 10$ and $\displaystyle BD=14$, making $\displaystyle AE = 34$. Neither scenario violates the conditions given.

Both statements are equivalent, as both are equivalent to stating that $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are collinear. Therefore, it suffices to determine whether the fact that the points are collinear is sufficient to answer the question. 

In both of the above figures,  $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are collinear, so the conditions of both statements are met. But in the top figure, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are the same ray, since $\displaystyle C$ is on $\displaystyle \overrightarrow{AB}$; in the bottom figure, since $\displaystyle B$ and $\displaystyle C$ are on opposite sides of $\displaystyle A$, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are opposite rays.

We show Statement 1 alone is insufficient to determine whether the two rays are the same by looking at the figures below. In the first figure, $\displaystyle B$ is the midpoint of $\displaystyle \overline{AC}$.

In both figures, $\displaystyle AC = 2 \cdot AB$. But only in the second figure, $\displaystyle B$ and $\displaystyle C$ are on the opposite side of the line from $\displaystyle A$, so only in the second figure, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are opposite rays.

Assume Statement 2 alone. If $\displaystyle B$ is the midpoint of $\displaystyle \overline{AC}$, then, as seen in the top figure, $\displaystyle B$ is on $\displaystyle \overrightarrow{AC}$. Therefore, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are the same ray, not opposite rays.

Statement 1 alone does not answer the question.

Case 1: Examine the figure below.

$\displaystyle AB+BC= AB + AB + AC > AC$,

thereby meeting the condition of Statement 1.

Also, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are opposite rays, since $\displaystyle B$ and $\displaystyle C$ are on opposite sides of the same line from $\displaystyle A$.

Case 2: Suppose $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are noncollinear. 

The three points are vertices of a triangle, and by the Triangle Inequality Theorem, 

$\displaystyle AB+BC > AC$.

Furthermore, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are not part of the same line and are not opposite rays.

Now assume Statement 2 alone. As can be seen in the diagram above, if $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are opposite rays, then by segment addition, $\displaystyle AB+BC = AC$, making Statement 2 false. Contrapositively, if Statement 2 holds, and $\displaystyle AB+ AC \ne BC$, then $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are not opposite rays.

We show Statement 1 alone is insufficient to determine whether the two rays are the same by looking at the figures below:

In both figures, $\displaystyle AB = 2 \cdot AC$, but only in the first figure, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are the same ray.

Assume Statement 2 alone. If $\displaystyle C$ is the midpoint of $\displaystyle \overline{AB}$, $\displaystyle C$ must be on $\displaystyle \overrightarrow{AB}$, as in the top figure, so $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are one and the same.

We show that both statements together provide insufficient information by giving two scenarios in which both statements are true.

Case 1: $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are noncollinear. The three points are vertices of a triangle, and by the Triangle Inequality Theorem, 

$\displaystyle AB+BC > AC$ and

$\displaystyle AB + AC > BC$.

Also, since the three points are not on a single line, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are parts of different lines and cannot be the same ray.

Case 2:  $\displaystyle \overline{AB}$ with length 2 and midpoint $\displaystyle C$.

$\displaystyle AB+BC =2+1 = 3$ and $\displaystyle AC = 1$, so $\displaystyle AB+BC > AC$; similarly, $\displaystyle AB + AC > BC$. Also, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are the same ray, since they have the same endpoint and $\displaystyle C$ is on $\displaystyle \overrightarrow{AB}$.

Statement 1 alone does not prove the rays to be the same or different, as seen in these diagrams:

In both figures, $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are collinear, satisfying the condition of Statement 1. But In the top figure, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are the same ray, since $\displaystyle C$ is on $\displaystyle \overrightarrow{AB}$; in the bottom figure, since $\displaystyle C$ is not on $\displaystyle \overrightarrow{AB}$, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are distinct rays.

Assume Statement 2 alone. Suppose $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are not the same ray. Then one of two things happens:

Case 1: $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are noncollinear. The three points are vertices of a triangle, and by the triangle inequality, 

$\displaystyle AB + BC > AC$,

contradicting Statement 2.

Case 2: $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are collinear. $\displaystyle A$ must be between $\displaystyle B$ and $\displaystyle C$, as in the bottom figure, since if it were not, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ would be the same ray. By segment addition, 

$\displaystyle AB + AC = BC$

$\displaystyle AB + AC+AB = BC+AB$

$\displaystyle AB + BC = AC + 2 \cdot AB > AC$,

contradicting Statement 2.

By contradiction, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are the same ray.