Constructing a triangle given three concurrent cevians? Well, I've been taught how to construct triangles given the $3$ sides, the $3$ angles and etc. This question came up and the first thing I wondered was if the three altitudes (medians, concurrent$^\text {any}$ cevians in general) of a triangle are unique for a particular triangle.
I took a wild guess and I assumed Yes! Now assuming my guess is correct, I have the following questions

How can I construct a triangle given all three altitudes, medians or/and any three concurrent cevians, if it is possible?

N.B: If one looks closely at the order  in which I wrote the question(the cevians coming last), the altitudes and the medians are special cases of concurrent cevians with properties


*

*The altitudes form an angle of $90°$ with the sides each of them touch.

*The medians bisect the side each of them touch.


With these properties it will be easier to construct the equivalent triangles (which I still don't know how) but with just any concurrent cevians, what other unique property can be added to make the construction possible? For example the angle they make with the sides  they touch ($90°$ in the case of altitudes) or the ratio in which they divide the sides they touch ($1:1$ in the case of medians) or any other property for that matter.
EDIT
What André has shown below is a perfect example of three concurrent cevians forming two different triangles, thus given the lengths of three concurrent cevians, these cevians don't necessarily define a unique triangle. But also note that the altitudes of the equilateral triangle he defined are perpendicular to opposite sides while for the isosceles triangle, the altitude is, obviously also perpendicular to the opposite side with the remaining two cevians  form approximately an angle of $50°$ with each opposite sides. Also note that the altitudes of the equilateral triangle bisects the opposite sides and the altitude of the isosceles triangle bisects its opposite sides while the remaining two cevians divides the opposite sides, each in the ratio $1:8$.
Now, given these additional properties (like the ratio of "bisection" or the angle formed with opposite sides) of these cevians, do they form a unique triangle (I'm assuming yes on this one) and if yes, how can one construct that unique triangle with a pair of compasses, a ruler and a protractor?

I seem to recall reading that two of the familiar sets of cevians --@Henry's comment to OP suggests that I must be thinking of altitudes and medians-- are easily proven to determine a triangle; but the last --angle bisectors-- is tricky (but still determinative?). Or maybe I'm thinking of something else entirely. Edit: @Beni's answer suggests that I'm remembering properly.

Well, here's a coordinate proof for the case of altitudes:
Consider altitudes of length $a$, $b$, $c$ dropped from respective vertices $A$, $B$, $C$. Place $C$ at $(0,c)$, vertex $A$ at $(-p,0)$, and vertex $B$ at $(q,0)$ for non-negative $p$ and $q$.
The equations of lines $AC$ and $BC$ are
$$AC: \quad -\frac{x}{p}+\frac{y}{c}-1=0 \qquad\qquad BC: \frac{x}{q}+\frac{y}{c}-1=0$$
Since $a$ is the distance from $A$ to $BC$, and $b$ the distance from $B$ to $AC$,
$$a = \frac{|-\frac{p}{q}-1|}{\sqrt{\frac{1}{q^2}+\frac{1}{c^2}}}=\frac{c \left(p+q\right)}{\sqrt{q^2+c^2}} \qquad b = \frac{c(p+q)}{\sqrt{p^2+c^2}}$$
We can solve this system for $p$ and $q$, getting
$$
p = \frac{c\left(a^2b^2-b^2c^2+c^2a^2\right)}{d} \qquad 
q = \frac{c\left(a^2b^2+b^2c^2-c^2a^2\right)}{d}$$
where
$$d^2 = \left( a b + b c + c a \right)\left(-a b + b c + c a \right) \left( a b - b c + c a \right)\left(a b + b c - c a\right)$$
Thus, a triple of altitudes determines a triangle (up to symmetry).

And here's one for medians:
Consider medians $a$, $b$, $c$ from respective vertices $A$, $B$, $C$. Place vertex $A$ at $(-p,0)$, vertex B at $(p,0)$, and vertex $C$ at $(c \cos t, c \sin t)$.
It's straightforward to compute the distance from $A$ to the midpoint of $BC$, and from $B$ to the midpoint of $AC$:
$$\begin{align}
a^2 &= \left(-p-\frac{1}{2}(p+c\cos t)\right)^2+\left(0-\frac{1}{2}(c \sin t)\right)^2 = \frac{1}{4}\left(9p^2+c^2+6pc \cos t\right) \\
b^2 &= \frac{1}{4}\left(9p^2+c^2-6pc \cos t\right)
\end{align}$$
Thus, 
$$a^2 + b^2 = \frac{1}{2}\left( 9 p^2 + c^2 \right) \quad \implies \quad 9 p^2 = 2 a^2 + 2 b^2 - c^2$$
so that
$$4 a^2 = 2 a^2 + 2 b^2 + 6 p c \cos t \;\implies\; \cos^2 t = \frac{\left(a^2-b^2\right)^2}{c^2 \left( 2 a^2 + 2 b^2 - c^2 \right)}$$
giving us $p$ and $t$, and again determining a unique triangle (up to symmetry).

I'm not sure about the case of general cevians. I'm not even sure how one would articulate a general case.
Certainly, within a given triangle, a set of cevians corresponds to a triple of ratios $(\alpha,\beta,\gamma)$ with $\alpha\beta\gamma=1$. (That's Ceva's Theorem, after all.) So, perhaps you might ask, 

For such a triple of ratios, does a triple of cevians $(a,b,c)$ uniquely determine a triangle?

Note, however, that the question doesn't include either the altitude case or the angle bisector case. For instance, we don't fix $(\alpha,\beta,\gamma)$ as we consider the universe of triangles with altitudes $(a,b,c)$.
Nevertheless, we can answer this version of the question in the affirmative. The proof is only slightly-modified from the median case:
We take the vertices $A(-p,0)$, $B(\gamma \, p,0)$, $C(c \cos t,c\sin t )$, and define cevian endpoints $D$ (on $BC$), $E$ (on $CA$), $F$ (on $AB$) such that 
$$\alpha=\frac{|DC|}{|BD|} \qquad \beta=\frac{|EA|}{|AC|} \qquad \gamma=\frac{|FB|}{|AF|}$$
(Of course, $F$ is the origin.) A little coordinate algebra gives
$$\begin{align}
a^2 &= |AD|^2 =\frac{1}{1+\alpha^2} \left(p^2(1+\alpha+\alpha\gamma)^2+2pc(1+\alpha+\alpha\gamma)\cos t+c^2\right)\\
b^2 &= |BE|^2 =\frac{1}{1+\beta^2} \left(p^2(1+\gamma+\beta\gamma)^2-2pc(1+\gamma+\beta\gamma)\cos t+c^2\right)
\end{align}$$
Thus, we can eliminate $p\cos t$ from the system:
$$\frac{1+\gamma+\beta\gamma}{1+\beta^2} a^2 + \frac{1+\alpha+\alpha\gamma}{1+\alpha^2} b^2 = P p^2 + Q$$
for $P$ and $Q$ that I won't write down here. Clearly, there's a unique solution for non-negative $p$, which in turn gives a solution for $t$.

As for geometric construction ... The toolset "a pair of compasses, a ruler, and a protractor" seems haphazardly suggested: If you have a ruler and protractor --assuming infinite precision-- then you don't need a compass at all to draw a straight-line figure.
Classic straightedge-and-compass construction isn't always possible, as one could easily take $(a,b,c)$ to be a non-constructible trio of lengths --for instance, take $a$ to define the unit length, and have $b=\pi$ and $c=\sqrt[3]{2}$-- and/or similarly take the ratios $(\alpha, \beta, \gamma)$ to ensure non-constructible segments.
