Well, I've been taught how to construct triangles given the $3$ sides, the $3$ angles and etc. This [question](https://math.stackexchange.com/questions/225236/how-to-find-the-area-of-a-triangle-with-lengths-of-heights) 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](https://math.stackexchange.com/users/6312/andre-nicolas) 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?