$\let\eps\varepsilon$This is not a complete answer. I will just show that $\theta$ can be smaller than $\pi/3$, but $\theta>\pi/4$.

**1.** Take an isosceles triangle $XYZ$ with $\angle Y=\angle Z=\pi/3+2\eps$ for a sufficiently small $\eps$. Cut out a small isosceles triangle $XY'Z'$. The resulting trapezoid $YZZ'Y'$ is a counterexample if $\eps$ and $XY'$ are chosen appropriately.

Namely, if $\eps$ and $XY'$ are small enough (the latter needs not be *very* small), then there are four halving chords with locally minimal length, joining $Y'Y$ with $YZ$, $Z'Z$ with $YZ$ (these two are equal), $Y'Z'$ with $YZ$ (this one is much larger if $XY'$ is small), and $YY'$ with $ZZ'$ (this one is larger than the first ones if $XY'$ is not too small comparable with $\eps$). Thus the first two are the shortest ones, and they subtend $\pi/3-\eps$.

Right now I cannot say how far this example can be extended. Also, it seems to be almost obvious that it is better to make $Y'Z'$ some curve rather than a line segment.

**2.** Now assume that the shortest halving segment of length $a$ joins $X$ and $Y$ and subtends equal angles $\theta<\pi/4$ at both its endpoints $X$ and $Y$. (It is easy to see that the shortest halving segment should connect two points different from vertices, and the subtended angles are equal). The supporting lines at $X$ and $Y$, together with $XY$, bound a triangle $\Delta$ of area $<a^2/4$, and a piece of $\partial C$ on its side of $XY$ lies inside $\Delta$.

Consider a ray perpendicular to $XY$, and move its endpoint along this piece of $\partial C$; at some moment it will contain a segment $I$ halving the area. The length of $I$ is at least $a$, the part of $I$ inside $\Delta$ is less than $a/2$, so its part outside $\Delta$ is larger than $a/2$. But then the convex hull of this part, together with $XY$, already makes an area $>a^2/4$, and lies in $C$. This contradicts the assumption that $XY$ is area-halving.

This proof works also for $\theta=\pi/4$, as the only polygon for which we get equalities everywhere is a square, but then it contains a halving segment with less length.

Again, this proof can be improved, but I do not know right now how far.

Citeseer link.) They do not address your question directly, but perhaps their proof methods can be applied. $\endgroup$ – Joseph O'Rourke Mar 27 '15 at 11:34