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In this diagram $AB=BC=CD=DE=1$meter,

    enter image description here

Prove that $\angle AEB+\angle ADB = \angle ACB$

The method I used involved simple calculations using the sine rule and the pythagorean theorem, but I used a calculator, which isn't the most intuitive means of proving the statement, especially since the answer at the back of the book gave the following as a proof:

By the sine rule: $$\sin(180-\angle ADB - \angle AEB)=\frac{5\left(\frac{1}{\sqrt5}\right)}{\sqrt{10}}=\frac{1}{\sqrt2}$$ $$180-\angle ADB - \angle AEB = 135^{\circ}$$ $$\implies\angle ADB + \angle AEB = 45^{\circ}$$ and $$\sin(\angle ACB)=\frac{1}{\sqrt2}$$ $$\angle ACB = 45^{\circ}$$

I feel like there is something very obvious that I'm overlooking, but I have yet to find it. My main area of confusion is how does $\displaystyle \sin(180-\angle ADB - \angle AEB)=\frac{5\left(\frac{1}{\sqrt5}\right)}{\sqrt{10}}$.

Any help is appreciated, thank you.

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we have $$\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\frac{1}{2}}=1=\tan(45^{\circ})$$

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    $\begingroup$ How does this clear up the area where my confusion lies? $\endgroup$ – joshuaheckroodt Nov 4 '17 at 18:26
  • $\begingroup$ you must use the tan-function, not sin $\endgroup$ – Dr. Sonnhard Graubner Nov 4 '17 at 18:27
  • $\begingroup$ I believe that, if OP were asking for a solution to the puzzle, then the question should be considered a duplicate of this one (and your solution effectively duplicates this answer). However, OP's specific concern is how a particular sine relation arises from the diagram. (Note: Your comment "you must use the tan-function" is an opinion apparently not shared by the author of OP's book.) $\endgroup$ – Blue Nov 4 '17 at 18:30
  • $\begingroup$ i don't know the author of OP's book $\endgroup$ – Dr. Sonnhard Graubner Nov 4 '17 at 18:31
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    $\begingroup$ @Dr.SonnhardGraubner: Of course you don't know the author. However, the solution in the book uses sine, and this is the source of the confusion for OP. Answers to the question should address that confusion, not offer alternative approaches to the puzzle (since these, as I mention, appear elsewhere on MSE). $\endgroup$ – Blue Nov 4 '17 at 18:33
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I had thought that "sine rule" meant "Law of Sines", but perhaps not. Consider this derivation of the relation at issue:

$$\begin{align} \sin(180^\circ-\angle ADB-\angle AEB) &= \sin(\angle ADB+\angle AEB) \tag{1}\\[4pt] &= \sin \angle ADB \cos \angle AEB + \cos \angle ADB \sin \angle AEB \tag{2}\\[4pt] &= \frac{|\overline{AB}|}{|\overline{AD}|}\frac{|\overline{AB}|}{|\overline{AE}|}+\frac{|\overline{BD}|}{|\overline{AD}|}\frac{|\overline{BE}|}{|\overline{AE}|} \tag{3}\\[4pt] &= \frac{1}{\sqrt{5}}\frac{3}{\sqrt{10}} + \frac{2}{\sqrt{5}} \frac{1}{\sqrt{10}} \tag{4}\\[4pt] &= \frac{5}{\sqrt{5}}\frac{1}{\sqrt{10}} \tag{5}\\[4pt] &= \frac{1}{\sqrt{2}} \tag{6} \end{align}$$ These steps are consistent with the appearance of "$5/\sqrt{5}$" in the book's proof, so it may be what the author intended. Even so, going from the sine expression to Step $(5)$ is quite a conceptual leap. Moreover, as shown in some answers to the puzzle as asked on MSE in 2016 (and in Dr. Graubner's solution here), a tangent-based derivation seems cleaner. (Of course, I believe a purely geometric approach such as the one I provided is cleaner still, but I'm biased.) All in all, it appears that the author could've done a much better job with the explanation.


I'll note that this puzzle appears on my Trigonography site, which includes a reference to both Numberphile's 2014 treatment, and Martin Gardner's 1996 Math Horizons article (which itself references a 1971 article by Charles Trigg(!), who found 54(!!) proofs of the relation).

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