About finding a point on a line AB that meets a known criteria Here is a problem that I am struggling with:

*

*Find a point C on line AB so that $AB^2 + BC^2 = 2
.AC^2$
I started by using Proposition 7 from Book II from Euclid's elements:
$AB^2 + BC^2 = AC^2 + 2.AB.BC$
For the left side to also be equal to $2.AC^2$ as required by the problem, it implies that:
$AC^2 = 2.AB.BC$
I know how to use propostion 11 in Book II to solve for finding a C such that:
$AC^2 = AB.BC$ but I could not figure out how to find a C that would enable $AC^2 = 2.AB.BC$
 A: Let point A be the origin, $A(0,0)$, and let point B be $B(b,0)$, so you want to find the coordinate of point $C(c,0)$
$$AB^2+BC^2=2AC^2\Rightarrow b^2+(c-b)^2=2c^2$$
Solve this quadratic equation and you get the coordinate of point C:
$$c=(-1+\sqrt{3})b,~~\text{or}~~c=(-1-\sqrt{3})b$$
A: First construct a square with a side $AB$ and call it $ABDB'$, where point $D$ is diagonally opposite to $A$. Construct $C$ between $AB$ with the given condition
$$AB^2 + BC^2 = 2AC^2$$
And a similar point $C'$ on the adjacent side $AB'$, symmetric about the diagonal $AD$.

By the Pythagorean theorem, the LHS becomes
$$AB^2+BC^2 = BD^2 + BC^2 = CD^2$$
And the RHS becomes
$$2AC^2 = AC^2 + AC'^2 = CC'^2$$
Together with the diagonal symmetry that $CD=C'D$, $\triangle CDC'$ is an equilateral triangle inscribed inside square $ABDB'$.
Then construct points on a square grid around point $A$, as shown in the diagram above. Both the horizontal and vertical distances between grid line have distance $AB$.
By considering the angles inside right angle $\angle D$ and the diagonal symmetry, $\angle B'DC' = \frac{90^\circ-60^\circ}{2} = 15^\circ$. And so $\angle B'C'D = 90^\circ-15^\circ = 75^\circ$. Then by symmetry about vertical $AB'$, $\angle B'C'E = \angle B'C'D = 75^\circ$.
By considering all the angles at point $C'$,
$$\begin{align*}
\angle EC'C &= 360^\circ - \angle EC'B' - \angle B'C'D - \angle DC'C\\
&= 360^\circ - 75^\circ - 75^\circ - 60^\circ\\
&= 150^\circ
\end{align*}$$
Since $EC'=EC'$, $C'D=C'C$ and $\angle EC'D = \angle EC'C = 150^\circ$, so $\triangle EC'D \cong \triangle EC'C$ (SAS). This gives the remaining side of $\triangle EC'C$
$$EC = ED = 2AB$$
By the Pythagorean theorem on right-angled $\triangle EFC$ (actually a $30-60-90$ triangle),
$$FC = \sqrt{EC^2-EF^2} = \sqrt{2^2-1^2}AB =\sqrt3AB$$
And the required point $C$ is at
$$\begin{align*}
AC &= FC-FA = \left(\sqrt3 -1\right) AB\\
BC &= FB-FC = \left(2-\sqrt3\right)AB
\end{align*}$$
