The eleven postulates are sufficient to prove 3.11.
Lemma 1 A line and a point not on it, two different lines in a plane, or two parallel lines define a plane.
Two points on a line and a point not on it define a plane by #7. If two lines are different there's a point on the second that's not on the first (by #6), so by the first part they define a plane. By definition two parallel lines are different lines in a plane so define it by the second part.
Lemma 2 If $a,b,t$ are different coplanar lines and $a$ is parallel to $b$ and $t$ is not parallel to $a$ then $t$ is a transversal of $a$ and $b$.
By definition $t$ intersects $a$ so call the point of intersection $A$ defining an angle $\angle at\ne 0$ (by #3). Let $S$ be a point on $b$ then $SA$ defines a line $s$ (by #6) which is a transversal of $a$ and $b$ (by definition). Then $s$ cuts off angles $\angle sb=\angle sa$ (by #10) and $\angle st\ne \angle sa$ (by #4 because they are coincident), so $t$ is not parallel to $b$ by $\angle st\ne \angle sb$ and #10, and is a transversal (by definition).
Proposition If $a,b,c$ are different lines with $a$ parallel to $b$ and $b$ parallel to $c$ then $a$ is parallel to $c$.
If the lines are coplanar then let $t$ be a line intersecting $b$, then applying Lemma 2 twice it is a common traversal of $a,b,c$. By #10 $\angle ta=\angle tb=\angle tc$ and by #11 $a$ is parallel to $c$.
If the lines are not coplanar, then let $C$ be a point on $c$. By Lemma 1 $a$ and $b$ are in a plane $\pi_1$, $b$ and $c$ are in a different plane $\pi_2$, and $a$ and $C$ are in a plane $\pi_3$. By #9 $\pi_2$ and $\pi_3$ intersect in a line $l$ that contains $C$.
$l$ cannot intersect $b$ in any point $B$, otherwise $a$ and $B$ are in both $\pi_1$ and $\pi_3$, so $\pi_1\equiv\pi_3$ by Lemma 1, $b\equiv \pi_1\cap\pi_2\equiv\pi_3\cap\pi_2\equiv l$ which would require $C$ to be on $b$, contradicting that $b$ and $c$ are parallel. So $l$ does not intersect $b$ but it intersects $c$ at $C$. Since $b,c,l$ are coplanar in $\pi_2$, by Lemma 2 they cannot all be different, so $l\equiv c$.
$l$ cannot intersect $a$ in any point $A$, otherwise $b$ and $A$ are in both $\pi_1$ and $\pi_2$, so $\pi_1\equiv\pi_2$ by Lemma 1, contradicting that $a,b,c$ are not coplanar. Since $l\equiv c$ and $a$ are both in $\pi_3$ and do not intersect it follows that $a$ is parallel to $c$.
This follows from the Intercept Theorem .
Let $DE$ be the line segment joining the midpoint $D$ of $AB$ and the midpoint $E$ of $BC$.
Draw a line parallel to to $DE$ that passes through $A$. Extend the side $BC$ so that it intersects this line in the point $F$. By the intercept theorem,
$$
{DB\over DA}={BE\over EF}
$$
But $DB=DA$, so, $EF=BE=EC$. It follows that $F=C$, and, thus, $AC$ is parallel to $DE$.
(Of course, you could argue using similar triangles too. The intercept Theorem is equivalent to "the similar triangle business".)
Best Answer
The trisection points of a side will be the two points which divide the side into thirds.
Hint: Consider the diagram below and the similar triangles: