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The inscribed angle theorem states that $\angle O=2\times\angle B$. The theorem is true for when point $B$ is located between points $A$ and $C$ relative to the perimeter. But what would happen if $B$ was located exactly at $A$ or $C$? Angle $B$ would obviously equal $0$ and so would the length of one of it's legs, but my question is, is this transition discrete or continuous? In other words, as the length of either line $BA$ or $BC$ approaches $0$, does angle $B$ also approach $0$? Or is it unaffected?

ShootinLemons
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1 Answers1

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First, I don't think you are correct when saying "$∠B$ would obviously equal 0" - this is like saying that an angle of a triangle that is incident with a side of length 0 is 0 while it is, by law of cosines, more accurately described as $arccos(\frac{1}{0})$ which is clearly undefined.

Second, a pre-condition of the theorem is that $AB$ and $BC$ are chords of the circle, which is not true if $B=A$ (or $B=C$). In all other cases, the theorem will hold. In particular, when $BA$ or $BC$ approaches 0, the angle at B will not approach 0 but stay equal to $\frac{∠O}{2}$, then "jump" to being undefined when $AB$ (or $BC$) reaches 0.

igel
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  • I think you are correct in saying that $arcos(B)$ would be undefined, but to say the angle itself wouldn't be zero is wrong. In your triangle analogy, a triangle with those properties wouldn't be a triangle. It would just be a line. So does that mean all lines are actually triangles with undefined angles? – ShootinLemons Oct 08 '21 at 10:59
  • It's a bit like saying all derivatives are undefined because the very definition of a derivative has a $0$ devisor in it. But that's clearly not the case, derivatives very much exist and are clearly defined. A simple example is the countless tangents you get when you take the derivative of a quadratic function at any input value. In other words, just because $arccos$ is undefined, does not mean $\angle B$ is undefined. It's an incorrect implication. $arcos$ is simply a representation of what value $\angle B$ holds, not it's definition. – ShootinLemons Oct 08 '21 at 11:08
  • Ah, maybe I misunderstood your question. If you want to know what happens when AB tends to 0 (as in $\lim\limits_{A\to B}$), then we can look at the proof on wiki (https://en.wikipedia.org/wiki/Inscribed_angle) and notice that all they need is that the sum of angles of an isosceles triangle with one side tending to 0 is still $\pi$. This is true since $\lim\limits_{x \to 0} x + 2*arccos(x^2/(2x)) = \pi$ (due to wolfram alpha). – igel Oct 11 '21 at 11:49