Construction of a Square in Clark's "Euclidean Geometry: A Guided Inquiry Approach" I am attempting to construct a square (Problem 58 in the title text) using a straight edge and compass and verify that what I have constructed is indeed a square (a rectangle with 4 congruent sides, where a rectangle is a polygon whose 4 segments meet at right angles). This is to be done in neutral geometry.
To begin with, I construct a line segment $AB$. I then construct perpendicular lines to $AB$ at $A$ and $B$ each. I then form $BC$ and $AD$ congruent to $AB$ on each of these lines, where $C$ and $D$ are on the same side of the line containing $AB$. From here, I am stuck. We are supposed to use what Clark calls Axiom 5: given a line $l$ and a point $P$ not on $l$, there is at most one line through $P$ parallel to $l$. If you combine this with a former result we have, you find that "at most" can be replaced with "exactly." I do not see how this helps me.
My ideas are as follows: form $CD$ and show it is congruent to and parallel to $AB$. I do not see why either of these should be true given what we know. The only ways we have developed so far to show two lines/segments are parallel involve showing the pair of lines/segments in question admit a transversal with either congruent alternating interior angles or congruent corresponding angles. The problem is we know nothing about the angles $BCD$ and $CDA$ a priori.
The other alternative is to construct a perpendicular segment $CE$ to $BC$ at $C$ and to somehow show that $D = E$, but my attempts at this (using combinations of SSS, SAS, AAS, ASA, and various versions of "SSA for right triangles," all of which we know) have failed. It seems I am either not recalling or am short one tiny fact needed to conclude.
This is not homework, but prep for a course I am teaching in the fall.
 A: I used your observation that the only way we know lines are parallel is if we have alternate interior angles.

So let $\overline{AB}$ be a segment and construct a perpendicular line $\ell$ to $AB$ passing through B, and find a point C on $\ell$ so that $BC \cong AB$ (so, just as you started). Now through construct a line $m$ perpendicular to $\ell$ and passing through $C$. By alternate interior angles, we know $m$ is parallel to line $AB$. Similarly, let $n$ be the line through $A$ perpendicular to $AB$, and so by alternate interior angles it is parallel to $\ell$.
First, we claim that $m$ and $n$ intersect. If not, then $m$ and $n$ are parallel and so the lines $\ell$ and $m$ are TWO different lines passing through $C$ that are parallel to $n$, contradicting Axiom 5. So let $D$ be the point of intersection of $m$ and $n$. The angle $ADC$ is right angle, again because of axiom 5 - draw the line $n'$ through $D$ perpendicular to $m$, by alternating interior angles $m$ must be parallel to $\ell$, and so $n, n'$ are two lines passing through $D$ parallel to $\ell$, so must be equal by axiom 5. So to summarize, so far we've created a rectangle with $AB\cong BC$.
I'm pretty sure you can figure it out from here, so I'll just outline one way:  draw $BD$, then apply Hypotenuse Leg, so then various angles that split a 90 degree angle are congruent and so they are 45, and then apply isoceles triangle theorem to get the sides of the rectangle are congruent. Let me know if you want me to add the details.
