Today I finished the 46 exercises from the review of analytic geometry, then rounded out my study time with more exercises from Analysis with an Introduction to Proof.
I had expected the proof from Revenge of the Squares, Part 1 to require showing that the figure must be a parallelogram. I managed it without that, but afterward I started thinking about how I would have proved that fact, had I needed it. The results are below.
Lemma: If, when two lines are intersected by another line, a pair of alternate interior angels is equal, then the first two lines are parallel.
Assume that two lines, $A$ and $B$, are not parallel.
Let them both be intersected by a third line, $C$. Since $A$ and $B$ are not parallel, they intersect, and $A$, $B$, and $C$ enclose a triangle.
Let $x$, $y$, and $z$ be the interior angles of the triangle, as shown in the diagram. Let the interior angle alternate to $x$ be called $k$ and the interior angle alternate to $y$ be called $j$.
Since the sum of the measures of the interior angles of a triangle is 180 degrees, $x+y+z=180$. Thus, $x=180-y-z$.
Furthermore, since $x$ and $j$ are complimentary supplementary, $x+j=180$, meaning that $x=180-j$.
Therefore, $$180-j=180-y-z$$ $$-j=-y-z$$ $$j=y+z$$
Since $z\neq 0$, it follows that $j\neq y$.
By similar reasoning, $k\neq x$.
Hence, if two lines are not parallel, then, when they are both intersected by a third line, neither pair of alternate interior angles is equal.
By contrapostition, it follows that if, when two lines are intersected by another line, a pair of alternate interior angels is equal, then the first two lines are parallel.
Proposition: A convex quadrilateral with each pair of opposite sides equal is a parallelogram (i.e. each pair of opposite sides is also parallel).
Consider a convex quadrilateral $ABCD$, where $AB=CD$ and $BC=AD$.
Draw $\overline{BD}$.
Since $AB=CD$, $BC=AD$, and $\overline{BD}$ is shared, $\triangle ABD\cong\triangle BCD$ by the side-side-side property.
Therefore, $\angle CBD\cong\angle ADB$ and $\angle BDC\cong\angle ABD$.
It follows, by the lemma, that $\overline{AB}\parallel\overline{CD}$ and $\overline{AD}\parallel\overline{BC}$.
Thus, $ABCD$ is a parallelogram.
SSS again. Proving that remains on the to-do list.
Fun and beautiful, as always. Just the quick comment that the lemma is basically the converse of the 5th Postulate, written in the complicated way Euclid does. And that by “complimentary,” you mean “supplementary.”
Thanks for the correction.
I proved the theorem first, using what later became the lemma, and then said “wait, is that an equivalence?” I knew the converse was at least very close to the parallel postulate, but wasn’t sure whether it held in this direction.