A Problem of Inscribed Equilateral 2-Polyline Given a regular strictly convex closed curve $\gamma:[0,1]\to\mathbb{R}^2,\ \gamma(0)=\gamma(1)$, prove or disprove that one can find three points $A, B, C\in\gamma$ such that $AB=BC$ and the area cut off by the chord $AB$ is equal to the area cut off by the chord $BC$ and they both are equal to the area between the chords (as it is depicted below).
I tried my best using twice Cauchy's intermediate value theorem, but does it really that easy? How can I prove or disprove the statement?

 A: The answer is YES. For such a regular convex closed curve, one can always find $A,B,C$ on it with $|AB| = |BC|$ and split the region it bounded into 3 sub-regions with equal areas.
The key is if one pick a point $E$ on the curve and locate another point $F$ to make chord $EF$ to cut out a fixed amount of area from the region, the $F$ so constructed depends on $E$ continuously.
Once one has that, the rest will be a simple application of IVT.
The real issue is how to justify $F$ depends on $E$ continuously.
Following is my clumsy attempt.

Given a convex body $K$ whose boundary $\partial K$ admits a $C^1$ parametrization with nonwhere vanishing derivatives.
Let $P, D$ and $\Delta$ be its perimeter, diameter and area.
Reparametrize $\partial K$ by arc length:
$$\gamma: [0,P] \ni s \mapsto \gamma(s) \in \partial K$$
Since $\partial K$ has at least one $C^1$ parametrization, $\gamma$ is also $C^1$. The condition of nowhere vanishing derivatives implies the tangent vector
$$\frac{d\gamma}{ds} = \left(\frac{dx}{ds},\frac{dy}{ds}\right) = (\cos\theta,\sin\theta)$$
is well defined and continuous. Up to modulo $2\pi$, so does the angle $\theta$.
WOLOG, we will assume orientation of $\gamma$ is positive. Since $K$ is convex, this means $\theta(s)$ is locally non-decreasing.  Extend parameterization $\gamma$ for all $s \in \mathbb{R}$ by
periodicity. The corresponding $\theta(s)$ can be lifted to a continuous non-decreasing function over $\mathbb{R}$.
For any $p = (p_1,p_2), q = (q_1,q_2) \in \mathbb{R}^2$, let $[p,q] = \frac12(p_1 q_2 - p_2q_1)$. When $p \ne q$, the pair $p, q$ defined an oriented line ($p \to q$). Let $H_{pq}$ be the closed half-plane consists of points on RHS of this line:
$$H_{pq} = \{ u \in \mathbb{R}^2, [ q-p, u ] \le 0 \}$$
For any $s \in \mathbb{R}$ and $t \in (s, s + P)$, let
$$K_{st} = K \cap H_{\gamma(s)\gamma(t)}
\quad\text{ and }\quad A(s,t) = \verb/Area/(K_{st})$$
Since $K_{st}$ is convex compact, $A(s,t)$ is supremum of area of all polygon inscribed within it. Since $\gamma$ is $C^1$, we can approximate $\gamma$ by a sequence of polylines, the convex hull of these polylines with $\gamma(s)$ allow us to generate a sequence of inscribed polygons which converges to $K_{st}$ (with respect to Hausdorff metric among convex bodies). We can evaluate the area of those inscribed polygons by shoelace formula. By passing to the limit, we obtain following formula for $A(s,t)$.
$$A(s,t) = \int_s^t [ \gamma(\tau) - \gamma(s), \gamma'(\tau) ] d\tau =
\int_s^t [ \gamma(\tau),\gamma'(\tau) ] d\tau - [\gamma(s),\gamma(t)]
$$
Since $\gamma$ is $C^1$, this formula tell us $A(s,t)$ is $C^1$ in both variables $s$ and $t$.
Since $K$ is convex, for $t_1, t_2 \in (s,s+L)$ with $t_1 < t_2$, we have
$$K_{st_1} \subset K_{st_2}\quad\implies\quad 
A(s,t_1) \le A(s,t_2)$$
This means $A(s,t)$ is non-decreasing in $t$.
In fact, we have a stronger result. For those $t$ where $A(s,t) \ne 0$ nor $\Delta$, we have
$$\partial_t A(s,t) \stackrel{def}{=} \frac{\partial A(s,t)}{\partial t} = [\gamma(t)-\gamma(s),\gamma'(t)]
= \frac12|\gamma(t)-\gamma(s)|\sin(\theta(s) - \phi)$$
where $\phi$ is the angle between $\gamma(t) - \gamma(s)$ and $x$-axis.
It is clear $|\gamma(t) - \gamma(s)| \ne 0$. Furthermore, $\sin(\theta(s) - \phi) \ne 0$ cannot vanish. Otherwise, the tangent line of $\gamma$ at $\gamma(t)$ will pass through $\gamma(s)$ and convexity of $K$ will lead to a contraction that $A(s,t)$ equals to either $0$ or $\Delta$.
Together with $A(s,t)$ is non-increasing in $t$, this leads to
$$A(s,t) \in (0,\Delta) \implies \partial_t A(s,t) > 0$$
From this, we can deduce there are $s \le t_a < t_b \le s + P$ such that
$$A(s,t)
\begin{cases}
= 0, & t \in (s,t_a]\\
\text{ strictly increasing }, & t \in (t_a,t_b)\\
= \Delta, & t \in [t_b,s+L)
\end{cases}$$
Notice for $t \in (s,s+\epsilon)$, we have the bound
$A(s,t) \le \pi \epsilon^2$. this means $A(s,t) \to 0$ as $t \to s+$.
By a similar argument, $A(s,t) \to \Delta$ as $t \to (s+L)-$.
By continuity of $A(s,t)$ in $t$, for any $\lambda \in (0,1)$, there exists $t \in (s,s+L)$ such that $A(s,t) = \lambda \Delta$. By above "increasing property" in $t$, we find such a $t$ is unique.
Let's call this $t$ as $q_\lambda(s)$.
For any fixed $\lambda \in (0,1)$ and $s_0 \in \mathbb{R}$, let $t_0 = q_\lambda(s_0)$.
Consider the graph in $\left\{ (s,t) \in \mathbb{R^2} : t - s \in (0,L) \right\}$ defined by an implicit equation:
$$F(s,t) \stackrel{def}{=} A(s,t) - \lambda \Delta = 0$$
This graph is non-empty as $(s_0,t_0)$ belongs to it. It is $C^1$ in both
$s$ and $t$. Finally, $\partial_t F(s,t) = \partial_t A(s,t)$ is non-zero at $(s,t) = (s_0,t_0)$. By Implicit function theorem, there is a  neighborhood $U = (s_0-\epsilon,s_0+\epsilon)$ of $s_0$ and a unique $C^1$ function $g : U \to \mathbb{R}$ with $g(s_0) = t_0$ and solves $F(s,g(s)) = 0$ on $U$.
Since $q_\lambda(s)$ satisfies last two conditions on $U$. It coincides with $g$ there. This means $q_\lambda(s)$ is $C^1$ over some neighborhood of $s_0$. Since $s_0$ can be arbitrary, $q_\lambda(s)$ is $C^1$ over $\mathbb{R}$.
By a similar construction for $t \in (s-L,s)$, we obtain the main lemma we need:

For any $\lambda \in (0,1)$, there exists two $C^1$ functions $p_\lambda$, $q_\lambda$ which are inverse to each other,
satisfy $s - P < p_\lambda(s) < s < q_\lambda(s) < s + P$ and
$$A(p_\lambda(s),s) = A(s,q_\lambda(s)) = \lambda \Delta$$

Let $\ell_p(s) = |\gamma(p_\lambda(s)) - \gamma(s)|$ and $\ell_q(s) = |\gamma(s) - \gamma(q_\lambda(s))|$. They are $C^1$ functions with period $P$. So $\ell_q(s)$ reaches absolute minimum at some $s_*$. Let $t_* = q_\lambda(s_*)$. Notice
$$\begin{align}
\ell_p(s_*) - \ell_q(s_*) &= \ell_q(p_\lambda(s_*)) - \ell_q(s_*) \ge 0\\
\ell_p(t_*) - \ell_q(t_*) &= \ell_q(s_*) - \ell_q(t_*) \le 0
\end{align}$$
The $C^1$ function $\ell_p(s) - \ell_q(s)$ changes sign over $(s_*,t_*)$.
By IVT, there is a $s^\dagger \in (s_*,t_*)$ such that
$$\ell_p(s^\dagger) = \ell_q(s^\dagger)
\quad\iff\quad
|\gamma(p_\lambda(s^\dagger)) - \gamma(s^\dagger)| = |\gamma(q_\lambda(s^\dagger)) - \gamma(s^\dagger)|
$$
Back to original problem, take $\lambda = \frac13$ and $\Delta = 3S$,
the three points $\gamma(q_\lambda(s^\dagger)), \gamma(s^\dagger), \gamma(p_\lambda(s^\dagger)$ will be the $A, B, C$ one need for the chords.
