Is this series representation of the hypotenuse symmetric with respect to the sides of a right triangle? This expression is more of a curiosity (perhaps even a tautology) than a practical method of finding the hypotenuse, as it requires taking the root of the sum of the squares of the catheti, which itself yields the result, ie, the length of the hypotenuse. Still, it leads to a question that might be of interest.
Given a right triangle with smaller cathetus a, larger cathetus b and hypotenuse c:
The infinite series is readily derived by constructing the normal line to the hypotenuse through the right angle and observing that this creates a similar triangle with dimensions in a ratio of b/c to the first. Iteration of this process ad infinitum  leads to a subdivision of the hypotenuse into lengths that converge to zero and whose infinite sum is equal to the hypotenuse.
My question is, as this series is derived with assumption that a is the smaller cathetus, is it symmetric with respect to both catheti, a and b? In other words, are both sides interchangeable in the series, and if so, why?
$$c=a^2\sum_{n=1}^\infty\frac{b^{2n-2}}{(a^2+b^2)^{\frac{2n-1}2}}$$

 A: Let us assume that the triangle is not degenerate, so that both $a$ and $b$ are non-zero.
Writing the series as
$$
 \frac{a^2}{\sqrt{a^2+b^2}} \sum_{n=0}^\infty \left( \frac{b^2}{a^2+b^2}\right)^n
$$
shows that this is a geometric series. It converges since $\frac{b^2}{a^2+b^2} < 1$. It is not required for the convergence that $a \le b$.
The value of the series is
$$
\frac{a^2}{\sqrt{a^2+b^2}} \cdot \frac{1}{1-b^2/(a^2+b^2)} = \sqrt{a^2+b^2} \, ,
$$
as expected. The same value is obtained if $a$ and $b$ are interchanged, since $\frac{a^2}{b^2+a^2} < 1$ as well.
A: The equality $\,\displaystyle c=a^2\sum_{n=1}^\infty\frac{b^{2n-2}}{(a^2+b^2)^{\frac{2n-1}2}}\,$ with $\,c = \sqrt{a^2+b^2}\,$ can be rewritten as:
$$
\frac{a^2}{a^2+b^2} \sum_{n=0}^\infty \left( \frac{b^2}{a^2+b^2}\right)^n = 1
$$
With $\displaystyle\,x = \frac{b^2}{a^2+b^2} \lt 1\,$ this reduces to the series expansion of $\displaystyle\,\frac{1}{1-x}\,$:
$$
(1-x) \, \sum_{n=0}^\infty x^n = 1
$$
The substitution $\,x \mapsto 1-x\,$ gives the symmetric form:
$$
x \, \sum_{n=0}^\infty (1-x)^n = 1 \;\;\;\;\iff\;\;\;\; \frac{b^2}{a^2+b^2} \sum_{n=0}^\infty \left( \frac{a^2}{a^2+b^2}\right)^n = 1
$$
In OP's question $\,x = \sin^2 B\,$, so the equality can also be written in trigonometric form as:
$$
\cos^2 B \, \sum_{n=0}^\infty \sin^{2n} B = 1
$$
The substitution $\,B \mapsto \dfrac{\pi}{2}-B\,$ gives the dual form:
$$
\sin^2 B \, \sum_{n=0}^\infty \cos^{2n} B = 1
$$
A: Your observation is that given a right triangle
in the plane, the line perpendicular to the
hypotenuse to the right angle partitions the
right triangle into two triangles similar to
the given one. Now suppose the larger subtriangle is chosen and now partition it as
in the original right triangle and continue the process to get the diagram in your question.
The hypotenuse is partitioned as you noticed, into line segments whose lengths form a geometric series whose sum is the length of the
hypotenuse.
However, suppose the smaller triangle is chosen instead. Then it forms a different hypotenuse
partition and another geometric series which
must have the same sum. Now suppose that the
partition process is performed on all of the
subtriangles which have a side along the
hypotenuse. Then the line segment partition
of the hypotenuse forms a doubly indexed
geometric series. Thus the symmetry of the
sum is made apparent. This new series is
$$ c\sum_{n=1}^\infty\sum_{m=1}^\infty \frac{b^{2n}a^{2m}}{c^{2n+2m}} =
\sum_{n=1}^\infty \frac{b^{2n}}{c^{2n-1}} \frac{a^2}{b^2} =
\sum_{m=1}^\infty \frac{a^{2m}}{c^{2m-1}} \frac{b^2}{a^2} = c.
$$
