Dividing a circular field in 4 equal parts by putting 3 equal straight fences In the book of Math Puzzles by Martin Gardner, an interesting way to divide a circular field in four equal parts by putting 3 equal curved (made up of circular arcs) fences has been discussed. The question here is: can one do the same by putting 3 equal straight fences.
Yesterday this question was discussed in MSE without the condition of equality of three straight fences.
You may see:
Dividing a circular field in 4 equal parts by putting 3 straight fences
 A: Yes, you can. Take a horizontal line that separates the top 25% of the circle from the rest of the circle, and call that the first fence. Reflect that line across the horizontal diameter of the circle to get the second fence. Finally, because we know the left most point of the first fence is a full diameter away from right most point of the second fence (by construction), we can say that there must exist some straight line that goes through the center of the circle that connects the first two fences together and is the same length as the other two fences because these other fences are smaller than the diameter of the circle. For visualization sake, the three fences together would look like a "Z" whose top and bottom line extends a bit further in both directions.
If you would like me to clarify anything please let me know :)
A: To integrate the preceding answer, in which the solution is given by two parallel equal chords (each identifying a segment equal to $1/4$ of the circle area) and an oblique third segment, we can calculate the length of the three fences by setting  
$$\frac{1}{2}R^2(\alpha-\sin{\alpha})=\frac{\pi}{4} R^2$$
where the LHS represents the area of a circular segment defined by an angle $\alpha$, expressed in radians. The numerical solution gives $\alpha=2.30988...$, corresponding to $\approx 132°$ degrees. The length of the chord is then 
$$ c=2R \sin\left(\frac{2.30988}{2}\right)  \approx  1.82954\, R$$
The same result can be obtained by integration. Considering a unitary circle of radius $R$ with the center in the origin, we can set
$$\int_a^1 \sqrt{1-x^2}  =\frac{\pi}{8} $$
Because the indefinite integral is $$  \frac{1}{2} \left(\sqrt{1 - x^2} x + \sin^{-1}(x)\right) +c$$
taking the definite integral and solving the equation we obtain the numerical solution $a=0.40397...$, which represents the distance of the first chord from the centre of the circle. The chord length is therefore
$$c= 2 \sqrt{1-a^2}\approx 1.82954...$$
After we have drawn two parallel equal chords of this length by tracing  vertical lines in $x=\pm a$, we can find the third segment (passing through the origin) by noting that, for a given angular coefficient $m$, the portion of the line $y=mx$ included between the two vertical lines has length $2\sqrt{a^2+(ma)^2}$. So setting
$$2\sqrt{a^2+(ma)^2}=c$$
and using the numerical values of $a$ and $c$ we get $m=2.03168...$, which means that the third segment crosses the two vertical chords at 
$$y=\pm  2.03168 \,a=\pm 0.82074...$$ Since the vertical chords crosses the circumference at $$y=\pm c/2 =\pm 0.91477...$$ as expected these crossing points are within the circle.
A: Following "Z" of @justaguy:
Let the ciecle be $x^2+y^2=r^2$ Let there be a horizontal chord AB, with $B(r\cos \theta, r \sin \theta)$. The length of the chord AB is $L=2r\cos \theta$. The area of the segmebt of the circle above AB is:
$$A=\frac{1}{2}(\pi-2\theta)r^2-\frac{r^2}{2} \sin 2\theta.$$
Let us demand it to be equal to the area quarter of the circle $\pi r^2/4$
This gives an implicit equation for $\theta$ as
$$(\pi-2\theta)-\sin 2 \theta=\pi/2.$$
We get the  root of this in $[0,\pi.2]$ numerically as $\theta=0.4158 \text{(radians)} = 23.8236^0 \implies L=1.8298 r~$
Let there be a point $P(h, r\sin \theta)$ on AB inside and near B. Similarly, by symmetry there will be a point $Q(-h,-r\sin\theta)$ on the chord CD. We demand $PQ=L$
this gives $h=r \sqrt{\cos 2\theta}$
Finally, for a given $r$, AB, CD and PQ are three equal straight line fences dividing the area of the circle in four equal parts. In this field th3re are two segments of the circles and other two equal parts are two "circular trapeziums": APQC and PBDQ, see 
the attached figure is for $r=1$.

A: It is possible if each of the three fences has one endpoint on the circumference, and the other endpoint somewhere along the length of another fence.  It is not possible if it is required to have both endpoints of each fence on the circumference.
First we will prove the second claim.  Suppose by contradiction that the three fences have their endpoints on the circumference.  Clearly, no two fences can intersect, because any two fences that intersect will divide the circle into four regions, and the third fence will divide the circle into at least one more region.  So the three fences are pairwise disjoint.  If the endpoints are labeled $(a_1, a_2)$, $(b_1, b_2)$, $(c_1, c_2)$ for segments $a, b, c$, a simple combinatorial argument shows that the order of the endpoints must be (up to a circular permutation or a reflection or a reversal of the indices within a line) one of two possibilities:  $(a_1, a_2, b_1, c_1, c_2, b_2)$, or $(a_1, a_2, b_1, b_2, c_1, c_2)$.  In the first case, a symmetry argument shows that segment $b$ must be a diameter, precluding $a$ and $c$ being diameters, thus this configuration is impossible.  In the second case, the dissection that maximizes each of the equal areas cut off by $a$, $b$, and $c$ is when $a$, $b$, $c$ form an equilateral inscribed triangle, and this does not allow for the central area to equal the three lune-shaped areas.  This concludes the proof by contradiction.
Now we turn our attention to a configuration that does work.  Draw an equilateral triangle $\triangle PQR$ with side length $x$ whose center coincides with the center of the circle, whose radius we may assume to be $1$ without loss of generality.  Extend $PQ$ to $Q'$ on the circumference so that $Q$ is between $P$ and $Q'$; similarly, extend $QR$ to $R'$ and $RP$ to $P'$.  This creates three congruent asymmetrical lunes.  We require the common area of these lunes to equal the area of the equilateral triangle; that is to say, we require $$ \frac{\sqrt{3}}{4} x^2 = \frac{1}{3} \left( \pi - \frac{\sqrt{3}}{4} x^2 \right),$$ or $$x = \frac{\sqrt{\pi}}{3^{1/4}} \approx 1.3467736870885982.$$

