An involute of a shortest path in a strictly convex surface in $\mathbb{R}^3$. Consider a smooth closed surface $\Sigma$ in $\mathbb{R}^3$ of positive Gaussian curvature, homeomorphic to a sphere. Here $U$ is the interior of the convex hull of $\Sigma$. When $c:[0,l]\rightarrow \Sigma$ is a minimizing geodesic of unit speed in it, then we defines $$a :[0,l]\rightarrow \mathbb{R}^3,\ a(t)=c(t)-t c'(t)$$ and $$f(t)=d(x,a(t))$$ for any fixed $x\in \Sigma$. Here $d$ is the intrinsic distance on $ \mathbb{R}^3- U$.

Problem : Then prove that $f(t)$ is increasing.

Thank you in advance.

Fix a point $x$ And if $c$ is a minimizing geodesic in $\Sigma$, we define $a(t)=c(t)-tc'(t)$ and $f(t)=d(x,a(t))$. We define $b$ s.t. $$\bigg\langle
 a(t+\varepsilon ),\frac{-c''(t)
}{|c''(t) |}  \bigg\rangle \frac{-c''(t) }{|c''(t) | } + a(t)=
b(\varepsilon )$$
If $\gamma=\gamma_\varepsilon$ is a shortest path from $x$ to
$b(\varepsilon)$, then assume that $$ \gamma(0)=x,\ \gamma(s)=y $$
and $$ y\in T_{c(t)}\Sigma \bigcap \gamma $$
Hence by Earnie060's answer, we have $$ f(t) = {\rm length}\ \gamma_0\leq {\rm length} \bigg(
\gamma|[0,s]\cap [y a(t)] \bigg)\leq {\rm length } \gamma $$
Hence $$\frac{d}{d\varepsilon }{\rm length}\ \gamma\geq 0 $$
Further note that
$$|f(t+\varepsilon )- {\rm length}\ \gamma |<C\varepsilon^2$$ for some $C>0$ and some $0<
\varepsilon <\delta$.
Hence $$ \frac{d}{d \varepsilon } f(t+\varepsilon ) = \frac{d}{dt}
{\rm length}\ \gamma $$
which completes the proof.
