A geometric reason why the square of the focal length of a hyperbola is equal to the sum of the squares of the axes?
Consider the hyperbola: $$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\tag{1} $$ Given $f_1,f_2$, the two foci of a hyperbola, one property of a hyperbola is that there is a constant, $\Delta\text{ distance}$, so that any point on the hyperbola, p, satisfies $$ \Big||p-f_1|-|p-f_2|\Big|=\Delta\text{ distance}\tag{2} $$ Consider a point at the intersection of the hyperbola and the line between the foci. The $\Delta\text{ distance}$ given in $(2)$ is the distance between the two branches of the hyperbola.
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Using $(1)$, we get that the distance between the two branches of the hyperbola is $2a$. Therefore, $$ \Delta\text{ distance}=2a\tag{3} $$ Consider a point at an infinite distance on the upper right branch of the hyperbola. The $\Delta\text{ distance}$ given in $(2)$ is $$ \Delta\text{ distance}=|f_1-f_2|\cos(\theta)\tag{4} $$ Using $(1)$, we get that $$ \begin{align} \tan(\theta)&=\lim_{x,y\to\infty}\frac yx=\frac ba\\ \cos(\theta)&=\frac{a}{\sqrt{a^2+b^2}}\tag{5} \end{align} $$ Combining $(3)$, $(4)$, and $(5)$, we get $$ |f_1-f_2|=2\sqrt{a^2+b^2}\tag{6} $$ Thus, if $c$ is the distance from the center of the hyperbola to each of the foci, then $(6)$ gives $$ c^2=a^2+b^2\tag{7} $$
Here's a solution using for the special case of a hyperbola formed as a "vertical" section of a cone ... that is, for the case in which the cutting plane is parallel to the cone axis.
Take a "sideways" look at one of the hyperbola's Dandelin Spheres; let $C$ be its center, and let it meet the cone tangentially at $T$. Our view of the configuration is parallel to the cutting plane (and perpendicular to the cone axis), which here appears as the vertical line $FM$. Were we to look perpendicular to the plane, point $M$ would be the center of the hyperbola, point $F$ would be a focus (so that $|MF| = |OC| = c$ in traditional notation), and point $V$ would be a vertex (so that $|MV| = a$); importantly ---and this is something of a leap of faith--- the crossed lines would be the asymptotes (with "slope" $\pm b/a$, where Dandelin radius $b = |CF|$ gives the rise for a run of $a$).
Since right triangles $\triangle OTC$ and $\triangle VMO$ have matching legs ($CT \cong OM$, via mutual congruence with $CF$) and acute angles ($\angle COT \cong \angle OVM$), the triangles are themselves congruent, so that $|OT|=|MV| = a$ and $$|OT|^2 + |TC|^2 = |OC|^2 \qquad \implies \qquad a^2 + b^2 = c^2$$
For now, I'll leave it as an exercise for the reader to tweak the argument to cover hyperbolas created by oblique cutting planes (as well as to cover ellipses), where the situation is much more complicated. The reader should also justify the "leap of faith" (which, of course, has to be so, because we know how the story ends).
BTW, the figure was drawn using GeoGebra.
Regarding the oblique case ... If the cone's generator makes an acute angle $\theta$ with its axis, and the cutting plane makes an acute angle $\phi$ with that axis, then one can show $$a = r \cos\theta \qquad c = r \cos\phi$$
where $r$ is the radius of the circle through the Dandelin centers, centered at their midpoint. Interestingly, the consequent relation $$b^2 = c^2 - a^2 = r^2 \left( \cos^2\phi - \cos^2\theta \right) = r^2 \left( \sin^2\theta - \sin^2\phi \right)$$ reveals $b$ as the geometric mean of the Dandelin radii, $r \left( \sin\theta - \sin\phi \right)$ and $r \left( \sin\theta + \sin \phi \right)$.
While it's possible to find $a$, $b$, $c$ represented by segments in a "sideways" diagram analogous to the one above, the connection between $a$, $b$, and the hyperbola's asymptotes is far less obvious. After all, the asymptotes arise from intersecting the cone, at its apex, with a plane parallel to the cutting plane; those lines don't have the same "slope" as the cone's generators, so the geometry of the sideways diagram isn't sufficient for capturing their nature.
You can solve this the same way as the ellipse, by taking the point with greatest y-value. However, that point lies at infinity, so instead of drawing an exact triangle, you take a limit of triangles with points on the curve (as you suggested) and scale them down so they don't go off to infinity themselves. This can be made precise in projective geometry, because then that point at infinity is a real point (there are 2, I believe). In fact, the hyperbola is just an ellipse turned inside out, so that it's greatest and least y-values lie at infinity (just like a line is a circle with a point at infinity).