Spherical cap area is $\pi r^2$. But why?

(If you're surprised by the title — $r$ is not what you (perhaps) think it is : )

Let $x$ be a point on a sphere $S$ and let $U$ be some sphere with center $x$ that intersects $S$.

Claim¹. The spherical cap cut out from $S$ and the circle cut out from $T_xS$ (tangent plane to $S$ at $x$) have the same area.

In other words, the area of a spherical cap is $\pi r^2$ where $r$ is the distance from its center $x$ to its boundary. (So we have a very simple formula for circle area in spherical geometry. But it's somewhat strange: $r$ is neither spherical nor Euclidean radius of the circle!)

Question. How to prove this geometrically?

It can be proved by direct computation² but surely there should be… some explanation why this map from a sphere to a plane is area-preserving, perhaps?

One interesting special case is $r=2R$ (where $R$ is the radius of $S$). The 'cap' in this case is the whole sphere — and $\pi r^2=4\pi R^2$. So a (good) answer to my question would give (yet another) explanation of sphere's surface area formula.

¹ I learned this from A. Akopyan.

² For example: the height of the cap is $r^2/2R$ (because $h/r=(r/2)/R$), so by Archimedes' hatbox lemma the area is $2\pi R\cdot r^2/2R=\pi r^2$.

I don’t know a simple geometric explanation for the total area, but infinitesimally, if the arc length is $a$ and the chord length is $r$, in the limit of small increments $da$ and $dr$, by similar triangles we have:

$$\frac{da}{dr}=\frac{r}{\rho}$$

where $\rho$ is the orthogonal distance of the endpoint of the arc/chord from a diameter that passes through the starting point. In spherical coordinates, $\rho = R \sin \theta$.

If the area of the spherical cap is $A$, we have:

$$\frac{dA}{da} = 2\pi \rho$$ $$\frac{dA}{dr} = \frac{dA}{da} \frac{da}{dr} = 2\pi r $$

Not a fully geometric proof of why the map preserves areas, but hope it helps.