A concise distance problem A falsely simple Euclidian geometry problem:

Points $A$, $B$, $C$ are collinear; $\|AB\|=\|BD\|=\|CD\|=1$; $\|AC\|=\|AD\|$.
  What is the set of possible $\|AC\|$ ?

I'm after a concise answer, with reasoning, that would get maximum points to an 11th-grader.
A related question asks an appropriate level for the problem (worded in less mathematical terms and thus reduced to distinct points).
To check your answer: the mean of the elements of the set of solutions to the present question is $\approx 1.08$.
 A: If $D$ was on the line $AB$, then either $D = A$, $B = C$, or $A, B, D, C$ would on a line (in the order). $D = A$ is impossible because that would imply $C = A$ and $||CD|| = 0$, and $B = C$ is impossible because that would imply $||AD|| = 2$but $||AC|| = 1$. The last case is impossible, because $||AC|| \neq ||AD||$.
Therefore, $D$ is not colinear with $A, B, C$. Additionally, $C \neq A$.
(edit: When $C = A$, $||CB|| = ||BD|| = ||DC|| = 1$, so an ᅟequilateral triangle $\triangle CBD$ can be formed.)
If $C$ was inside $AB$, since $D$ is on a circle with center $A$ and radius $||AC||$, projection of $D$ on $AB$ should be between $A$ and $C$. However, since $||BD|| = ||CD||$, projection of $D$ should be located at midpoint of $BC$. Therefore, it is impossible.
Therefore, either $C$ is located on [1] ray $AB$ - $AB$, or [2] ray $BA$ - $AB$.
The first case
$\angle BAD = \angle BDA$ since $||AB|| = ||BD||$. $\angle DBC = \angle DCB$ since $||BD|| = ||CD||$. Since $\angle DBC = 2 \angle DAB$, the sum of inner angles of triangle $ADC$ can be calculated by
$$\angle DAC + \angle ADC + \angle ACD = 5 \angle DAC = 180°$$
Therefore, $\angle DAC = 36°$. Now, since $\triangle ACD \sim \triangle DBC$,
$$||AC|| : ||CD|| \equiv ||DB|| : ||BC||$$
$$||AC|| : 1 \equiv 1 : ||AC|| - 1$$
And now $||AC||$ can be calculated by solving a quadratic equation and the fact that $||AC|| > 1$.
The second case
Do the same thing. But now $||BC|| = ||AC||+1$.
A: W.l.o.g. choose coordinates as follows:
\begin{align*}
A &= \begin{pmatrix}0\\0\end{pmatrix} &
B &= \begin{pmatrix}1\\0\end{pmatrix} &
C &= \begin{pmatrix}x\\0\end{pmatrix} &
D &= \begin{pmatrix}\frac{x+1}2\\
\sqrt{x^2-\left(\frac{x+1}{2}\right)^2}\end{pmatrix}
\end{align*}
Point $D$ is choosen on the perpendicular bisector of $BC$. This bisector is only uniquely defined and a strict requirement if $B\neq C$, so the case of $B=C$ will have to be handled separately. It is furthermore chosen at distance $x$ from $A$. So the above coordinates will already ensure the following conditions:
\begin{align*}
\lVert AB\rVert &= 1 &
\lVert BD\rVert &= \lVert CD\rVert &
\lVert AC\rVert &= \lVert AD\rVert
\end{align*}
Now all that remains is the condition $\lVert BD\rVert = 1$ or equivalently $\lVert CD\rVert = 1$. I go for the latter.
\begin{align*}
1 = \lVert CD\rVert^2 &=
\left(x-\frac{x+1}2\right)^2 + \left(\sqrt{x^2-\left(\frac{x+1}{2}\right)^2}\right)^2
\\&= \left(\frac{x-1}2\right)^2 + \left(x^2-\left(\frac{x+1}{2}\right)^2\right) \\
4 &= \left(x^2-2x+1\right) + \left(4x^2-x^2-2x-1\right) \\
0 &= 4x^2 - 4x - 4 = x^2 - x - 1 \\
x_{1,2} &= \frac{1\pm\sqrt{1+4}}{2} \\
x_1 &= \frac{1-\sqrt5}2 \approx -0.618 \\
x_2 &= \frac{1+\sqrt5}2 \approx 1.618
\end{align*}
The special case of $B=C$ gives a third solution $x_3=1$ describing $C$, with an associated point $D_3=\begin{pmatrix}\frac12\\\frac{\sqrt3}2\end{pmatrix}$ (although these coordinates of $D_3$ are not neccessary to answer the question). The requested length is the absolute value of $x$, so the final solution is
$$\lVert AC\rVert = \lvert x\rvert \in \left\{
\frac{\sqrt5-1}2, 1, \frac{\sqrt5+1}2 \right\}$$
The idea of choosing suitable coordinates without loss of generality will probably be the most problematic concept. Whether students will be able to think of this on their own, particularly in an exam situation, depends a lot on the details of their education. After that, chances are good that many will miss the special case, since the coordinates (as perhaps derived from a sketch of a non-degenerate situation) seem to perclude such a third solution. The computation of $x$ on the other hand should be pretty straight forward, once they got this far.
