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.


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.
