My back is somewhat better today. I spent an hour on exercises in Analysis with an Introduction to Proof and I plan to read some more of Infinite Powers this evening, as well.
I’m also ready to prove the proposition about inscribed angles from my post Groundwork. It turns out that proposition falls right out of another that reader Tim McL suggested starting with when I said I knew little about the geometry of circles.
Proposition: If a central angle and an inscribed angle are subtended by the same arc of a circle, then the measure of the central angle is twice that of the inscribed angle.
Let $A$ and $B$ be points on a circle with center $D$.
Draw the central angle subtended by arc $AB$.
Draw an inscribed angle also subtended by arc $AB$ and let its vertex be called $C$.
Join $\overline {CD}$.
Now suppose that the sides of the two angles intersect only at $A$ and $B$ (as in the figure above).
Since $\overline{BD}$ and $\overline{CD}$ are both radii of the circle, $BD=CD$, from which it follows that $\angle CBD = \angle BCD$ (as I’ve proven previously).
Similarly, $\overline{AD}$ and $\overline{CD}$ are both radii of the circle, so $AD=CD$ and $\angle CAD = \angle ACD$.
Futhermore, since the sum of the angles of a triangle is $\pi$ radians, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.
It is also true that $\angle BCD + \angle ADC + \angle ADB = 2\pi$.1
Combining these equations will yield $\angle ADB = 2(\angle BCD + \angle ACD)$.
But $\angle ACB = \angle BCD + \angle ACD$, so $\angle ADB = 2\angle ACB$, and the theorem is proven for this case.
Suppose instead that sides of the angles intersect other than at $A$ and $B$, as shown above. (Note that this argument does apply in the edge case where $\overline{BC}$ intersects $\overline{AD}$ at $D$.)
Then, $\angle ACB =\angle ACD-\angle BCD$ and $\angle ADB = \angle BDC – \angle ADC$.
As above, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.
Combining these equations will yield $\angle ADB = 2(\angle ACD – \angle BCD)$.
Hence $\angle ADB = 2\angle ACB$, and the theorem is proven for this case, as well.
After proving this proposition, I looked up how Euclid did it. His solution was largely the same. It was slightly more elegant, using the property of triangles illustrated below to cut out some of the algebra, but not so elegant as to dispense with the cases.
Proposition: If two inscribed angles are subtended by the same arc of a circle, then they are equal.
I’m just going to sketch a proof of this. By the proposition above, the blue and green inscribed angles subtended by arc $AB$ both have half the measure of the red central angle subtended by the same arc. Therefore, they must be equal to each other.
Footnotes
- Do I need to prove this? I have an idea of how I would do it, but it seem so fundamental. ↩︎
Um, It seemed to me that you did just prove it. I don’t see where it lacks generality, and it shows that <ADB=<ACB, what else would be in a proof?
That was a footnote to “It is also true that $\angle BCD + \angle ADC + \angle ADB = 2\pi$.” I agree it was kind of confusing. I’ve added a label.
Just to show that people really are reading and reflecting carefully on your thoughts, the line with the footnote 1 has a small typo. It should read $\angle BDC+\angle ADC+\angle ADB=2\pi$.
I agree that it seems pretty obvious that the sum of the angles around a point always adds up to $2\pi$, or, as Euclid would say, to 4 right angles. I don’t remember Euclid proving this fact, but he may. It’s a generalization of what he proves in I.13, that if you take a point $D$ on the straight line $BDC$ and pick any point $A$, the the sum of the angles $\angle BDA+\angle ADC$ equals 2 right angles. If Euclid doesn’t exactly prove your result, then it’s probably because, having gotten I.13, he figures he can always get your theorem whenever he needs it.
It’s nice that you and Euclid end up with the same cases. Euclid often has a tendency to leave out some cases, doing only the hardest ones. And in a way, your argument differs from his significantly because you have algebra and he doesn’t.
Thank you for the correction. I’ll see to it when I’m at my laptop tomorrow.