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.
