Elementary trigonometric proof problem using multiple angles in the sine rule 
In this diagram $AB=BC=CD=DE=1$meter,
    
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.
 A: we have $$\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\frac{1}{2}}=1=\tan(45^{\circ})$$
A: 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). 
A: 
I want to give a trigonometric and a geometric solution.
Here is the trigonometric solution.
Clearly $\alpha = m\angle BCA = 45^\circ$
So we need to show that
$\beta + \gamma = m\angle BDA + m\angle BEA = 45^\circ$
Clearly $0 < \beta, \gamma < 45^\circ$.
So $0 < \beta + \gamma < 90^\circ$.
So, if we can show that $\tan(\beta+\gamma) = \tan 45^\circ = 1$,
then it will follow that $\beta + \gamma = 45^\circ$
\begin{align}
   \tan(\beta+\gamma) 
   &= \dfrac{\tan \beta + \tan \gamma}{1 + \tan \beta \cdot \tan \gamma} \\
   &= \dfrac{\frac 12 + \frac 13}{1 - \frac 12 \cdot \frac 13} \\
   &= \dfrac{\frac 56}{1 - \frac 16} \\
   &= 1
\end{align}
Here is the geometric solution.
$\triangle ABE \sim \triangle AHG$
$\triangle ABD \sim \triangle GFA$
$m\angle HAF = 45^\circ = \alpha$
$\therefore \alpha + \beta = \gamma$.
