Prove that an isometry preserves straight lines using intuitive geometry This is an exercise of the book "Basic Mathematics" by Serge Lang, p.145.
I've been working on this proof for a few days and I can't seem to make it more coherent than this. Would appreciate some help.
Prove: An isometry preserves straight lines. In other words, if $L$ is a straight line and $F$ is an isometry, then $F(L)$ (the image of $L$ under $F$) is also a straight line. If $L$ is the line passing through two distinct points $P$, $Q$, then $F(L)$ is the line passing through $F(P)$ and $F(Q)$.
Proof: Let $F$ be an isometry. 
Let $P$ and $Q$ be distinct points on the plane through which a unique line $L$ passes through.
Let $M$ be a point on $L$ other than $P$ and $Q$.
For simplicity, we denote $P' = F(P)$, $Q' = F(Q)$ and $M' = F(M)$.
We can distinguish three possible cases as follows:
Case (i) $M$ lies on $\overline{PQ}$.
Since $F$ is an isometry, we have
$$d(P, M) = d(P', M')$$
$$d(M, Q) = d(M', Q')$$
Since $M$ lies on the segment between $P$ and $Q$
$$d(P, Q) = d(P, M) + d(M, Q)$$
$$d(P, Q) = d(P', M') + d(M', Q')$$
$$d(P', Q') = d(P', M') + d(M', Q')$$
Hence the images of points $P$, $Q$ and $M$ lie on a same straight line, under $F$.
We wish to show that every point of such line can be expressed as the image of a point on $L$ under $F$.
Let $M'$ be a point at distance $r$ from $P'$. Let $M$ be a point at distance $r$ from $P$.
Then $F(P) = P'$ must be at distance $r$ from $F(M)$. It follows that $F(M) = M'$.
Case (ii) Let $M$ be a point on the ray of vertex $P$, passing through $Q$, at a distance from $P$ greater than that of $Q$. Then
$$d(P, M) = d(P, Q) + d(Q, M)$$
$$d(P, M) = d(P', Q') + d(Q', M')$$
$$d(P, M) = d(P', M')$$
Therefore, the image of points $P$, $Q$, and $M$ under $F$ lie on a same straight line aswell.
By similar reasoning as before, we can show that every point of such line can be expressed as the image of $L$ under $F$.
Case (iii) Let $M$ be a point on the ray  of vertex at $Q$, passing through $P$ at a distance from $Q$ greater than that of $P$. This is similar to the previous case, with the difference that $P$ and $Q$ play opposite roles. So we have
$$d(Q, M) = d(Q, P) + d(P, M)$$
$$d(Q, M) = d(Q', P') + d(P', M')$$
$$d(Q, M) = d(Q', M')$$
The same conclusion can be drawn for this case.
We have thus proven that the image of a straight line $L$ passing through distinct points $P$ and $Q$ is the straight line passing through $F(P)$ and $F(Q)$. q.e.
 A: You definitely found the necessary argument to prove the result, but you seem having trouble sparing yourself some extra cases. The result you want to prove goes as follows. 
If $3$ points $P_1,P_2,P_3$ are colinear, then $F(P_1),F(P_2),F(P_3)$ are colinear. To see this, assuming (without loss of generality) that $P_2 \in \overline{P_1 P_3}$, we have 
$$
d(F(P_1),F(P_3)) = d(P_1,P_3) = d(P_1,P_2) + d(P_2,P_3) = d(F(P_1),F(P_2)) + d(F(P_2),F(P_3)),
$$
which implies the colinearity. Therefore, if $P,Q$ are any two points and $R$ is a third point on the line passing through $P$ and $Q$, the result implies that $F(R)$ lies on the line passing through $F(P)$ and $F(Q)$ (since the three points $F(P),F(Q),F(R)$ are colinear). 
Using this notion of "colinearity" avoids the apparition of cases. Your three cases essentially deal with switching the role of "who is $P_2$", but we are only interested in the colinearity of the three points, not on the name of the point lying between the two other ones.
This shows that $F(line) \subseteq line$. To prove equality, I assume you use techniques of affine geometry, but I'll leave that to you. You could also use the fact that the isometry will be invertible and the inverse is also an isometry, proving that $F^{-1}(line) \subseteq line$.
Hope that helps,
