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Is there a proof in (or from) Euclid's Elements of the following statement, or something akin to it?

Produce a circle with radius $OA$. Let $P$ be the point on that circle with an acute angle $\theta$ from $OA$ and let $B$ be the intersection of the straight line extending $OA$ with the tangent line to the circle at $P$. Then the length of the arc $PA$ is smaller than the length of the segment $PB$.

Analytically speaking, if $O$ is the origin and $A = (1,0)$ then this statement amounts to saying that $\theta < \tan(\theta)$. I can give an analytic proof of this by comparing the areas of the triangle $OPB$ (area $= \frac{1}{2} (\sin(\theta)^2 \tan(\theta) + \sin(\theta) \cos(\theta)) = \frac{1}{2}\tan(\theta)$) and the sector $OPA$ (area = $\frac{1}{2}\theta$) but I was wondering if there was a purely geometric proof. Maybe a not-so-straightforward application of Common Notion 5?

Lorenzo Riva
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You can see that $[\triangle OAB] -$ (the area of sector OAP)=$\frac{1}{2}(\tan\theta - \theta) > 0$ by geometry. This directly leads to the comparison of length of the arc and the segment.

Hypernova
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  • What I'm missing is very last part. I can see how that would follow if $AP$ were a line segment. What guarantees I can compare arcs with segments? – Lorenzo Riva Jan 02 '23 at 16:48
  • @LorenzoRiva I compared areas to get $\tan\theta > \theta$, and applied this fact back to lengths. It doesn't directly compare arcs and segments. – Hypernova Jan 02 '23 at 16:50
  • Right, so in the non-analytic formulation we can compare the area of $\triangle OPB$ and the area of the sector $OPA$ and notice that the difference is positive. From there, how can one infer that the difference of the sides in question is positive? – Lorenzo Riva Jan 02 '23 at 16:52
  • How do you show that the ratio of the length of the subtended arc to the radius is the same as twice the ratio of the area of the sector to the square of its radius? – eyeballfrog Jan 02 '23 at 16:55
  • @eyeballfrog I just used the formula that I know, which is that the area of the sector OAP is $\frac{1}{2}\cdot 1 \cdot \theta=\frac{1}{2}\theta$, and also the length of arc $AP$ is $\theta$. Maybe this can be some circular logic. – Hypernova Jan 02 '23 at 16:59
  • We can just use that $2\pi : \pi r^2 = \theta : S$. I am not sure if we need a proof of this. – Hypernova Jan 02 '23 at 17:05
  • @Hypernova I guess what I was after is exactly a proof of this. Some quick googling shows that Proposition XII.2 (http://aleph0.clarku.edu/~djoyce/elements/bookXII/propXII2.html) has $A = 4C r^2$ for some constant $C$, where $A$ is the area of a circle with radius $r$, and according to https://math.stackexchange.com/questions/542870/did-euclid-prove-that-pi-is-constant Euclid did not write anything about the area of a sector. – Lorenzo Riva Jan 02 '23 at 17:17
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    We MUST use a definition of the numerical length of a circular arc, as the limit of a suitable sequence of approximations by sets of line segments, e.g. $\theta =\lim_{n\to\infty}2^n\tan (\theta /2^n).$ This is a strictly decreasing sequence, and the term with $n=0$ is just $\tan\theta.$ – DanielWainfleet Jan 02 '23 at 17:25