Is it possible to construct three cevians so that three of the parts are equivalent? Recently I have started wondering if it is possible to construct two cevians in a triangle so that at least three of four parts which the cevians divide the triangle in are equivalent. I think the answer is yes, but do not know how to construct them.
 A: In case of an isosceles triangle the things are clear, it is possible in the one or the other way. Let us consider the general remained case.
Let $\Delta ABC$ be a triangle with sides of different lengths.
Let $BE$ and $CD$ be the two cevians, let $X$ be their intersection.
(1)
We consider first the case when two of the three "equivalent pieces" are the two triangles $\Delta XBD$ and $\Delta XCE$.
There is the same angle measure in $X$, opposite angles, so the products $XB\cdot XD$ and $XC\cdot XD$ are equal, we are then forming proportions equivalently instead, and get the similitude  $\Delta XBC\sim\Delta XED$, so $DE$ is parallel to $BC$!
So let $DE$ be a "moving" parallel to $BC$.
Let $x$ be the proportion $x=AD:AB=AE:AC$, $x\in[0,1]$. Let $S$ be the surface (i.e. area) of $\Delta ABC$. Then
$$
\begin{aligned}
\operatorname{Area}(ABC) &= S\ ,\\[2mm]
\operatorname{Area}(ABE) &= xS\ ,\\
\operatorname{Area}(BEC) &= (1-x)S\ ,\\[2mm]
\operatorname{Area}(ACD) &= xS\ ,\\
\operatorname{Area}(BDC) &= (1-x)S\ ,\\[2mm]
x&=
\frac{DE}{BC}
=
\frac{XD}{XC}
=
\frac{XE}{XB}
\\
\frac x{1+x}
&=
\frac{XD}{CD}
=
\frac{XE}{BE}
\\[2mm]
\operatorname{Area}(XBD) &=
\frac{XD}{CD}\cdot\operatorname{Area}(BCD) = 
\frac x{1+x}\cdot (1-x)S\ ,\\
\operatorname{Area}(XCE) &=
\frac{XE}{BE}\cdot\operatorname{Area}(BCE) = 
\frac x{1+x}\cdot (1-x)S=\operatorname{Area}(XBD)\ ,
\\[2mm]
\operatorname{Area}(XBC) &=
\frac{XC}{CD}\cdot\operatorname{Area}(BCD) = 
\frac 1{1+x}\cdot (1-x)S\ ,
\\
\operatorname{Area}(ADXE) 
&=\left(
1 - 2\cdot\frac{x(1-x)}{1+x}-\frac{1-x}{1+x}
\right)S
=\frac{2x^2}{1+x}S\ .
\end{aligned}
$$
So one has to choose $x$ so that either
$$
2x^2 = x(1-x)\qquad\text{ or }\qquad 1-x =x(1-x)\ .
$$
The first case leads to the solution $x=1/3$, after simplifying with $x$. (We avoid degenerated cases.)
So the winning configuration is

(Note: The above picture could have been the answer, but i do not like such haiku solutions, not even in novel style. I really like the full story. But well, for the novel style it would have been enough to mention that the median in the picture separates the triangle in two having the same area, and than the side proportions $1:2$ reflect the proportion of the triangles with one vertex in $X$ and opposite sides the corresponding $1:2$ parts...) 
(2) The remained case is when the two triangles different from $\Delta XBC$ are not both on the list of the three equivalent pieces. Let us eliminate the piece $\Delta XCE$. Then $X$ is the mid point of $CD$ to make the triangles $\Delta XBC$ and $\Delta XBD$ have the same area. We denote by $x$  the proportion 
$$ x=\frac{AD}{AB}\ .$$
Then from the theorem of Menelaos, in triangle $\Delta ADC$, w.r.t. the line $EXB$,
$$
\frac{EA}{EC}\cdot
\frac{XC}{XD}\cdot
\frac{BE}{BA} =1
\ ,
$$
we get corrected, thanks @greenturtle3141, $EA:EC=1:(1-x)$. From here, denoting again by $S$ the total area:
$$
\begin{aligned}
\operatorname{Area}(ABC) &= S\ ,\\[2mm]
\operatorname{Area}(DBC) &= (1-x)S\ ,\\[2mm]
\operatorname{Area}(XBD) &= \frac 12(1-x)S\ ,\\
\operatorname{Area}(XBC) &= \frac 12(1-x)S\ ,\\[2mm]
\operatorname{Area}(ABE) &= \frac {1}{1+(1-x)}S\ ,
\end{aligned}
$$
so we write the equation
$$
\frac {1}{2-x}S=2\cdot\frac 12(1-x)S\ ,
$$
and the solution in $(0,1)$ of the equation $(2-x)(1-x)=1$, i.e. $x^2-3x+1=0$, is
$$
\frac 12(3-\sqrt5)\approx
0.381966011250105\dots\ .
$$
The better way to see it is take the well known golden $1-x$,
$$
1-x=\frac 12(\sqrt 5-1)\approx 0.61803398874989\dots\ ,
$$
(I tried to find a good geometric reason why this is a solution... still search for the lighting picture, but i have to submit the error correction now...)
A: 
Yes you can!  In the triangle above, our cevians are $BE$ and $CF$.  The three areas that are equal are $[BFX]$, $[AFXE]$, and $[CEX]$.
The point $X$ is defined with barycentric coordinates $(1/2, 1/4, 1/4)$.  In other words, we have the ratios:
$$\frac{DX}{DA} = \frac12$$
$$\frac{EX}{EB} = \frac14$$
$$\frac{FX}{FC} = \frac14$$
There are actually two other solutions, and they involve some other regions.  If you're looking for just one example, then this is the one.
First, I will show that the triangle above satisfies your condition, with three of these areas being equal.  Since $\frac{EX}{EB} = \frac{FX}{FC}$, we see that $\triangle XEF \sim \triangle XBC$, so $EF || BC$.  This implies that the heights of $\triangle FBC$ and $\triangle EBC$ are the same, hence their areas are equal.  This implies that $[FBX] = [ECX]$.
Now, let both of these areas be $A$.  We see that $[BCX] = 3A$ by base ratios, and hence, $[BDX] = [CDX] = 1.5A$.  Since $[BXD] = [BXA]$, we must have $[AFX] = .5A$, and similarly, $[AEX] = .5A$, so $[AFXE] = A$, as desired.  This result can be derived using easier methods, like barycentric coordinates and mass points. 
Now I'll derive the other two possibilities.  First, I prove that the quadrilateral $AFXE$ must be one of the three equal area regions.  Assume that it is not.  Then $[BFX] = [BXC] = [CEX]$, and thus, $FX = XC$ and $EX = XB$.  But the triangles $BFX$ and $CEX$ are directly congruent and $BF||CE$, which is absurd.
Now, let $X$ have homogenized barycentric coordinates $(x, y, z)$ so we have $x+y+z=1$.  We then have $F(x:y:0) = (\frac{x}{x+y}, \frac{y}{x+y}, 0)$, and trivially, $A(1,0,0)$.  If we assume without loss of generality that $[ABC]=1$, then the area of $\triangle AFX$ can be evaluated with the following determinant:
$$[AFX] = \begin{vmatrix}
1&0&0\\
\frac{x}{x+y}&\frac{y}{x+y}&0\\
x&y&z\\
\end{vmatrix}=\frac{1}{x}\cdot\frac{1}{x+y}\cdot\begin{vmatrix}
x&0&0\\
x&y&0\\
x&y&z\\
\end{vmatrix}=\frac{xyz}{x(x+y)} = \frac{yz}{x+y}$$
We slew the determinant by factoring out common factors in rows, and then using row reduction to make the computation trivial.
We use the same method to derive the following areas:
$$[AFXE]  =\frac{yz}{x+y}+\frac{yz}{x+z}$$
$$[BFX]  =\frac{xz}{x+y}$$
$$[CEX]  =\frac{xy}{x+z}$$
$$[BXC]  = x$$
The last area can be derived from interpreting barycentric coordinates as areal coordinates.
With this new algebraic foundation, we will now completely eliminate $[BXC]$ as one of the possible areas.  We know that $[AFXE]$ is one of the areas.  First, without loss of generality, let $[BFX]$ be another one of the possible areas.  Then:
$$\frac{yz}{x+y}+\frac{yz}{x+z} = \frac{xz}{x+y}$$
This expands and simplifies to:
\begin{equation}
(*)\ \ \ y^2 + 2xy + yz = x^2 + xz
\end{equation}
Now assume that $[BXC]$ is one of the areas.  Then:
$$[BFX] = [BXC] \implies \frac{xz}{x+y} = x \implies x+y=z$$
Since $x+y+z=1$, We derive that $x+y=z=\frac12$.  Hence $y = \frac12-x$, so we can eliminate $y$ and $z$ in $(*)$:
$$(\frac12-x)^2 + 2x(\frac12-x) + (\frac12-x)(\frac12) = x^2+\frac12x$$
This has the unique positive root $x = \frac{\sqrt{5}-1}{4}$.  Hence we have the solution:
$$x = \frac{\sqrt{5}-1}{4}$$
$$y = \frac{3-\sqrt{5}}{4}$$
$$z = \frac12$$
Similarly, if we chose $[CEX]$ to instead be our second area, we would have the solution:
$$x = \frac{\sqrt{5}-1}{4}$$
$$y = \frac12$$
$$z = \frac{3-\sqrt{5}}{4}$$
This wraps up the case of $[CEX]$ being a possible area.  Our last case to consider $[AFXE] = [BFX] = [CEX]$, which will result in the solution I presented first.
$(*)$ still holds true, and now we have:
$$\frac{xz}{x+y} = \frac{xy}{x+z}$$
This expands to:
$$xy+y^2 = xz + z^2$$
Now if we restate $(*)$ for the reader's convenience:
$$y^2 + 2xy + yz = x^2 + xz$$
We see that the equations can be subtracted to obtain:
$$xy + yz = x^2 - z^2$$
Factoring:
$$y(x+z) = (x+z)(x-z) \implies x = y+z$$
Once again, from $x+y+z=1$, this implies $y+z=x=\frac12$.  Substitute $z = \frac12 - y$ into $(*)$ to solve for $y$, so we derive the following solution:
$$x = \frac12$$
$$y = \frac14$$
$$z = \frac14$$
Which is precisely the original solution.
Ok this might have been slightly overkill.
