As Joseph O'Rourke said, curve-shortening flow can be used to prove the theorem of the three geodesics. If a simple closed curve on a Riemannian surface evolves by curve-shortening flow and does not collapse to a point in finite time, then it smoothly approaches a geodesic. This is a result of Grayson.
Curve-shortening flow can also be used in a really beautiful way to prove the isoperimetric inequality in two dimensions for closed embedded curves. This works more generally on Riemannian surfaces, but it's really neat and tidy for $\mathbb{R}^2$. Edit: Gage originally showed that the quantity $\frac{L^2}{4\pi A}$ is non-increasing with respect to this flow.
Consider a closed embedded initial curve $\gamma_0$ (which we give a positive orientation), and let $\gamma_t$ be the solution to curve-shortening flow. By the Gage-Hamilton-Grayson theorem, $\gamma_t$ remains embedded, is smooth, and shrinks to a round point in finite time, i.e. the curve becomes rounder as it shrinks to a point and "converges" to a circle in a suitable sense. This is truly a remarkable theorem if you look at the spiral posted by Joseph O'Rourke--it's not obvious that such a thing should unravel and become convex. It's also remarkable considering how finnicky and prone to singularities mean curvature flow and Ricci flow are (in higher dimensions at least).
Now, we have the following evolution equations for curve-shortening flow for the length $L$, area $A$, and curvature $k$ of the curve:
$$\frac{\partial L}{\partial t} = - \int_{\gamma_t} k^2\, ds$$
$$\frac{\partial A}{\partial t} = -\int_{\gamma_t} k\, ds = -2\pi$$
The integrals are taken with respect to arc length, and the second equality in the second line follows from the fact that we are considering a positively-oriented simple closed curve so its winding number is 1. The first equation shows that the length is non-increasing, and in fact curve-shortening flow can be thought of as the flow of curves that decreases length "fastest" or "most efficiently". Using the above evolution equations, we can compute
$$\frac{\partial}{\partial t}(L^2 - 4\pi A) = -2L \int_{\gamma_t} k^2\, ds + 4\pi \int_{\gamma_t} k\, ds$$
$$= -2L \int_{\gamma_t} k^2\, ds + 2(2\pi) \int_{\gamma_t} k\, ds$$
$$=-2L \int_{\gamma_t} k^2\, ds + 2 \big(\int_{\gamma_t} k\, ds\big)^2$$
$$\leq -2L \int_{\gamma_t} k^2\, ds +2\big(\int_{\gamma_t} \, ds\big)\big(\int_{\gamma_t} k^2\, ds \big) $$
$$= -2L \int_{\gamma_t} k^2\, ds + 2L \int_{\gamma_t} k^2\, ds = 0$$
Thus, $L^2 - 4\pi A$ is non-increasing in time with respect to curve-shortening flow. As we mentioned before, closed embedded curves shrink to round points, so $L^2 \to 4\pi A$, since $L^2 = 4\pi A$ for circles. So, $L^2 - 4\pi A \to 0$. However, since $L^2 - 4\pi A$ is non-increasing in time, this means that for $\gamma_0$, $L^2 > 4\pi A$. This is the isoperimetric inequality in two dimensions for closed embedded curves.
Also, note that we may conclude that $L^2 = 4\pi A$ if and only if the curve is a circle. In the above calculation, we only applied Cauchy-Schwarz which is sharp exactly when $k$ is constant. For a closed embedded curve, this can only be a circle, by the fact that the curvature determines a curve up to isometry (fundamental theorem of planar curves).
