Q:

Geometric construction to divide a segment

Given a segment AB, I would like to construct using only straightedge and compass, a point C on the segment AB such that $\frac{AC}{CB}$ is equal to $\frac{\phi}{2}$, where $\phi$ is the golden ratio, $\phi = 1.61803..$ .
The Wikipedia article on the golden ratio offers a construction for dividing a segment in the ratio $\phi$, but I cannot figure out how to do it to divide in $\phi/2.$ Any ideas?

A:

Render
$\frac{1+\phi /2}{\phi /2}=\sqrt{5}$
So the entire length divided by the shorter piece is $\sqrt{5}$.
Start with the given segment $AB$.  Bisect this segment to identify the midpoint $M$ and construct a circle $Z$ centered at $M$, passing through $A$ and thus also through $B$.  Construct the perpendicular to $AB$ through $A$, and mark off a point $C$ on this perpendicar, such that $AC$ is congruent to $AM$.  Draw the hypoteneuse $BC$ which intersects circle $Z$ at point $X$ (distinct from $B$).  Triangle $XBA$ is similar to triangle $ABC$ so $AB$ is $\sqrt{5}$ times as long as $AX$.  Construct $Y$ on $AB$ with $AY$ congruent to $AX$ and render $Y$ as the required dividing point.

