Explaining this proof without words for $\frac{\pi}{4} = 2 \arctan\frac{1}{3} + \arctan\frac{1}{7}$

Question

Proof of Hutton's 1776 formula:

Why does the figure establish this identity?

$$ \frac{\pi}{4} = 2\arctan\left(\frac{1}{3}\right) + \arctan\left(\frac{1}{7}\right) $$

Answer 1

Let $\alpha$ be the angle $GAB$, let $\beta$ be the angle $HAB$, and let $\gamma$ be the angle $CAB$.

Notice that $\alpha+\beta+\gamma=\frac{\pi}{4}$ since the triangle $ACD$ is right and $\tan(A)=5/5$.

Notice also that the angles $AHC$ and $AHG$ are $\frac{\pi}{2}$ since $AH$ bisects the line segment $CG$ and the lengths of $AC$ and $AG$ are equal.

With the Pythagorean theorem and basic trigonometry definitions, $\tan\alpha=\frac17$, $\tan\beta=\frac{\sqrt 5}{3\sqrt 5}$, and $\tan\gamma=\frac{\sqrt 5}{3\sqrt 5}$. Hence $\alpha=\arctan(1/7)$ and $\beta=\gamma=\arctan(1/3)$. It follows that

$$\frac{\pi}{4}=\alpha+\beta+\gamma=2\arctan\frac13+\arctan\frac17.$$

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Answer 2

I would like to attract your attention on the fact that this relationship, here interpreted under a geometrical setting, can also be interpreted in a very easy way under a purely algebraic setting using complex numbers.

Indeed, consider identity :

$$50(1+i)=(3+i)^2(7+i)\tag{1}$$

Dividing (1) by $50 \sqrt{2}$ gives complex numbers with modulus $1$ :

$$\frac{(1+i)}{\sqrt{2}}=\left(\frac{(3+i)}{\sqrt{10}}\right)^2\frac{(7+i)}{\sqrt{50}}\tag{2}$$

Now, you get the desired relationship by equating arguments on each side of (2), knowing that, due to the fact that all angles are in $[0, \pi/4]$ :

$$\begin{cases}\arg(z_1z_2)&=&\arg(z_1)+\arg(z_2)\\ \arg(a+ib)&=&\arctan \frac{b}{a}\end{cases}\tag{3}$$

Remark : relationship (1) can be considered in the ring of "Gaussian integers" (the set of complex numbers with integer real and imaginary parts).