I intended to split my study time today between analytic geometry review and logic review, but ended up getting drawn into some exploration as well. The result is that I have two proofs about quadrilaterals ready to post. The first of the two appears below. It’s a proof of my hunch from the post Squares Again that if a convex, four-sided, equilateral polygon has one right angle, then it has four right angles. It turns out that I could prove this without knowing that such a figure must be a parallelogram. (It must, though. That’s the proof for tomorrow.)
Lemma: The angles opposite (i.e. subtended by) the equal sides of an isosceles triangle are also equal.
Consider the isosceles triangle $\triangle ABC$ with $AB=AC$.
Draw a line bisecting $\angle CAB$ and extend it to intersect $\overline{BC}$. Let the point of intersection be called $D$.
Now, by the side-angle-side property, $\triangle ACD\cong\triangle ABD$, since $AB=AC$, $\angle CAD\cong\angle BAD$, and $\overline{AD}$ is shared.
Therefore, $\angle ACD\cong\angle ABD$. These are the angles opposite the equal sides of the original isosceles triangle, so the lemma is proven.
Proposition: If a convex, four-sided, equilateral polygon has one (interior) right angle, then it has four (interior) right angles.
Let $ABCD$ be an convex, four-sided, equilateral polygon, and let $\angle BCD$ be a right angle.
Divide the polygon by drawing $\overline{BD}$.
Note that, since $\overline{BD}$ is shared and all other line segments are equal, $\triangle ABD\cong\triangle BCD$ by the side-side-side property.
Hence, since $\angle BCD$ is a right angle and the angles of congruent triangles are equal, $\angle BAD$ is also a right angle.
Similarly, $\angle ADB\cong\angle CBD$.
Furthermore, by the lemma, since $\triangle ABD$ is isosceles, $\angle ADB\cong\angle ABD$, and since $\triangle ACD$ is isosceles, $\angle BDC\cong\angle CBD$.
Therefore, $\angle ADB\cong\angle ABD\cong\angle BDC\cong\angle CBD$.
Let the measure of those angles be called $x$. Since the sum of the measures of the angles of a triangle is 180 degrees, either triangle gives us the relationship $90+2x=180$, so $x=45$.
The measures of $\angle ADC$ and $\angle ABC$ are both $2x=2(45)=90$. Thus they are also right angles.
Hence, if a convex, four-sided, equilateral polygon has one right angle, then it has four right angles.
This proof uses SSS again—I actually thought of it while trying to work out a proof of SSS—as well as SAS, through the lemma1. It also uses the fact that the measures of the angles of a triangle sum to 180 degrees. I’m pretty sure that theorem is only a couple of steps from the parallel postulate, though.
My original version of this proof didn’t make use of the lemma about isosceles triangles, but instead argued that the two congruent triangles could be mapped onto each other by either rotation or reflection, and that was why all of the acute angles had to be equal. That was kind of a nifty argument, but not easy to formalize.
Stay tuned tomorrow for the proof about parallelograms.
- To do. ↩︎
I love it that this work continues to draw you in.
So do I. Hooray!
and that you continue to draw, in this work.
Heh.
Wonderful, as always!
May I make a few Euclidean comments?
(1) This lemma is part of Book I, Prop. 5 in Euclid, so it’s the first thing after SAS.
(2) Euclid can’t prove it your way, because Euclid only allows himself to work with things he can construct. He hasn’t yet shown how to make perpendicular lines with compass and straightedge, so he can’t do your first step. If you like, you can look at the Elements to see how he gets around this.
(3) The argument I’ve always wanted to use to prove the lemma, which may be the sort of thing you allude to at the end, is this: Triangles BAC and CAB are congruent by SAS, so angle B equals angle C. This is somehow not an argument Euclid would allow, but I still feel like it’s the proof from the Book, as Erdős would say. Of course, that assumes that figures are congruent to their reflections.
(4) Euclid doesn’t prove that the sum of the angles of a triangle is 2 right angles until later. Partly that’s because he tries for a long time not to use the 5th Postulate. It’s as if he’s aware of non-Euclidean geometry, despite being Euclid and all. In any case, he does neutral geometry without the 5th Postulate until I.27, I think.
One way to see this is that in I.16 he proves that an exterior angle of a triangle is greater than each of the interior and opposite angles. Similarly, I.17 shows that the sum of any 2 angles of a triangle is less than 2 right angles. Then in I.32, he has started to use the 5th Postulate, and he proves that the exterior angle is not just greater than either of the opposite interior angles. It is their sum. He also proves in I.32 that not only do 2 angles of a triangle add up to less than 2 right angles, but all 3 angles add up to 2 right angles.
If Euclid thought that there was nothing in the Fifth Postulate that made him uncomfortable, then there wouldn’t have been any reason to prove I.16 and I.17. He would just have waited and proved I.32. So he must have been asking himself how far he could without the 5th Postulate. He could get I.16 and I.17, but to get all the way to I.32, to the sum of the angles of a triangle being 2 right angles, he needed the 5th Postulate.
Euclid was way subtler than people give him credit for who just say, “Well, by 1900 people were being more careful with the axiomatics.”
(2) $\overline{AD}$ is an angle bisector, not a perpendicular. I find that the angle bisector construction is Proposition 9 of Book 1, though, so if the lemma is covered in Proposition 5, Euclid still can’t have proven it the way I did.
(4) Cool. I did not know that there was evidence that Euclid was interested in doing without the parallel postulate. I guess he might have thought that it ought to be provable from the others, as many later mathematicians did.
I often think about the summary of Euclid’s postulates that I heard you give once. The one about drawing a line segment means, “I have a straight edge.” The one about drawing a circle means, “I have a compass.” The one about extending a line segment indefinitely means, “Space is infinite.” The one about all right angles being congruent means, “Space is uniform.” And the parallel postulate means, “Space is flat.” I find that very useful both for understanding their significance and for remembering them.
What nice phrasings. I must once have been thoughtful. It’s delightful that one of us remembers those things. You see, you have always been a mathematician. And, as the other comments say, an artist. I, too, love all the figures.