Today I spent all my study time on exploration and proofs, as opposed review exercises. My first goal was to prove Theorem 2 from yesterday:
Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1m_2=-1$; that is, their slopes are negative reciprocals:$$m_2=-\frac{1}{m_1}$$
I succeeded in that, and also came up with an argument for the converse of the Pythagorean Theorem, which is needed for the proof.
Argument for the converse of the Pythagorean Theorem
Proposition: Given a triangle with side lengths $a$, $b$, and $c$, if $a^2+b^2=c^2$, then the triangle is a right triangle.
Let $\triangle ABC$ be a triangle such that $CA=\sqrt{AB^2+BC^2}$.
Construct a right triangle $\triangle EFG$ such that $EF=AB$ and $FG=BC$.
By the Pythegorean Theorem, $GE=\sqrt{EF^2+FG^2}$. Furthermore, $\sqrt{EF^2+FG^2}=\sqrt{AB^2+BC^2}=AC$.
Thus, $EF=AB$, $FG=BC$, and $GE=AC$, meaning that by the side-side-side property, $\triangle EFG$ is congruent to $\triangle ABC$.
The angles of congruent triangles are equal. Therefore, since $\triangle EFG$ is a right triangle, $\triangle ABC$ is also a right triangle.
The use of the SSS property here arose from a hint I found on the Internet. I would like to be able to prove that property, but have had no luck so far.1 It isn’t obvious to me why it should be true that having all sides equal guarantees congruence for triangles when it does not for other polygons.
An abbreviated proof of Theorem 2
Proposition: Two lines with slopes $m$ and $n$ are perpendicular if and only if $mn=-1$; that is, their slopes are negative reciprocals:$$n=-\frac{1}{m}$$
Let $y=mx+b$ and $y=nx+c$ be two lines such that $m\neq n$.
By Theorem 1 (from yesterday), since $m\neq n$, the two lines are not parallel. Thus, they intersect.
Let $E$ be the point at which the lines intersect.
We know from the proof of Theorem 1 that the coordinates of $E$ are $\left(\frac{c-b}{m-n},\frac{mc-nb}{m-n}\right)$.
Let $F\neq E$ be point on $y=mx+b$ with coordinates $(x_F, y_F)$ and let $G\neq E$ be point on $y=nx+c$ with coordinates $(x_G, y_G)$.
Furthermore, let $x_F=x_G=\frac{c-b}{m-n}-1$.
By plugging $\frac{c-b}{m-n}-1$ into both equations, we can find the values of $y_F$ and $y_G$ in terms of $m$, $n$, $b$, and $c$.
Using the formula below, we can then find the distance between each pair of points:
$$P_1P_2=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$ where $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$.
The algebra yields the following: $FG=|m-n|$, $EF=\sqrt{1+m^2}$, and $EG=\sqrt{1+n^2}$.
Now, assume that the lines $y=mx+b$ and $y=nx+c$ are perpendicular. This means that they meet at a right angle. Thus $\triangle EFG$ is a right triangle, and by the Pythagorean Theorem
$$\left(\sqrt{1+m^2}\right)^2+\left(\sqrt{1+n^2}\right)^2=\left(|m-n|\right)^2$$$$m^2+n^2+2=m^2-2mn+n^2$$$$2=-2mn$$$$mn=-1$$
Therefore, if the two lines, $y=mx+b$ and $y=nx+c$, are perpendicular, then $mn=-1$.
Instead, assume that $mn=-1$. In that case, $$\left(\sqrt{1+m^2}\right)^2+\left(\sqrt{1+n^2}\right)^2=m^2+n^2-2mn=(m-n)^2\text{.}$$
It follows, by the converse of the Pythagorean Theorem, that $\triangle EFG$ is a right triangle. Since the right angle of a right triangle is opposite the hypotenuse, $\angle FEG$ is a right angle. Hence the lines $y=mx+b$ and $y=nx+c$ are perpendicular.
Therefore, if the slopes of two lines, $y=mx+b$ and $y=nx+c$, have a product of $-1$, then the lines are perpendicular.
I had some other ideas today, as well, including regarding the question about squares from my post Squares Again. Expect those in the next few days.
