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I am trying to prove the inscribed angle theorem using vectors. I fixed the dots $A=(\cos\theta,\sin\theta)$, and $B=(\cos\varphi,\sin\varphi)$, and I took another point $C=(\cos\psi,\sin\psi)$ in the biggest arc $AB$.

My idea was to calculate $\dfrac{\langle A-C,B-C\rangle}{\lvert{A-B}\rvert\lvert{B-C\rvert}}$, what according to my calculations is $$\dfrac{1+\cos(\theta-\varphi)-\cos(\psi-\theta)-\cos(\varphi-\psi)}{2\sqrt{1+\cos(\theta-\psi)\cos(\varphi-\psi)-\cos(\theta-\psi)-\cos(\varphi-\psi)}}.$$ My main difficulty here is the square root, which I can't get rid of. Does someone know how to proceed from here?
Or maybe to solve the problem with vectors with a different approach?

Lorenzo Andreaus
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    You're mostly there. If you simplify the expression, you'll get something that doesn't depend upon $\psi$, which tells you what you want to know. With regard to simplifying the numerator, try re-writing the first two terms together, then also the last two terms. Inside the square root in the denominator, combine the first and third terms, and then also the second and fourth. That should get you going. – Blue Aug 03 '20 at 03:35
  • @Lorenzo did you aim to write |A-C||B-C| in the denominator? – user376343 Aug 04 '20 at 07:19
  • I didn't know exactly what to expect. I was just simplifying the expression and hoping something was going to cancel. – Lorenzo Andreaus Sep 01 '20 at 23:55

1 Answers1

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You expanded your terms all the way. Take a few steps back: \begin{align} \langle A-C, B-C\rangle &= \langle (\cos\theta-\cos\psi,\sin\theta-\sin\psi),(\cos\phi-\cos\psi,\sin\phi-\sin\psi)\rangle\\ &= (\cos\theta-\cos\psi)(\cos\phi-\cos\psi)+(\sin\theta-\sin\psi)(\sin\phi-\sin\psi)\\ &= -2\sin\frac{\theta-\psi}2\sin\frac{\theta+\psi}2 \cdot(-2)\sin\frac{\phi-\psi}2\sin\frac{\phi+\psi}2 \\&\quad+ 2\cos\frac{\theta+\psi}2\sin\frac{\theta-\psi}2 \cdot 2\cos\frac{\phi+\psi}2\sin\frac{\phi-\psi}2\\ &= 4\sin\frac{\theta-\psi}2\sin\frac{\phi-\psi}2\left(\sin\frac{\theta+\psi}2\sin\frac{\phi+\psi}2 + \cos\frac{\theta+\psi}2\cos\frac{\phi+\psi}2\right)\\ &= 4\sin\frac{\theta-\psi}2\sin\frac{\phi-\psi}2\cos\left(\frac{\theta+\psi}2-\frac{\phi+\psi}2\right)\\ &= 4\sin\frac{\theta-\psi}2\sin\frac{\phi-\psi}2\cos\frac{\theta-\phi}2 \end{align}

\begin{align} |A-C|^2|B-C|^2 &= ((\cos\theta-\cos\psi)^2+(\sin\theta-\sin\psi)^2)((\cos\phi-\cos\psi)^2+(\sin\phi-\sin\psi)^2)\\ &= (2-2\cos\theta\cos\psi - 2\sin\theta\sin\psi)(2-2\cos\phi\cos\psi - 2\sin\phi\sin\psi)\\ &= 4(1-\cos(\theta-\psi))(1-\cos(\phi-\psi))\\ &= 4\cdot 2\sin^2\frac{\theta-\psi}2\cdot 2\sin^2\frac{\phi-\psi}2\\ &= 16 \sin^2\frac{\theta-\psi}2\sin^2\frac{\phi-\psi}2 \end{align}

and hence $$\cos\measuredangle ACB = \frac{\langle A-C, B-C\rangle}{|A-C||B-C|} = \cos\frac{\theta-\phi}2$$ which means that the angle $\measuredangle ACB$ does not depend on $C$ and is equal to half the central angle: $$\measuredangle ACB = \frac12(\theta-\phi) = \frac12 \measuredangle AOB.$$

mechanodroid
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