Can the median, angle bisector and the altitude of a triangle intersect to form an equilateral triangle? On a sheet of paper, a blue triangle is drawn. A median, a bisector and an altitude of this triangle (not necessarily from three distinct vertices) are drawn red. The triangle dissects into several parts. Is it possible that one of these parts is an equilateral triangle with red sides?
I tried to get the angles of the triangle formed, but failed because I was not able to find the angle formed by the median.
 A:  
I'll consider the case where altitude, angle bisector, and median are drawn from different vertices, and go at it in reverse, i.e. start with an equilateral triangle $ABC$. 
On $CB$ extended erect a perpendicular at any point $D$, meeting $CA$ extended at $E$. At $E$ reflect $ED$ about $EC$, meeting $BC$ extended at $F$. Let $AB$ extended meet $ED$ at $G$, and join $FG$.
Thus within triangle $EFG$ we have an equilateral triangle $ABC$ with sides $BC$ and $CA$ lying on altitude $FD$ and angle bisector $EJ$.  If side $AB$ lies on the median, then $H$ is the midpoint of $EF$. 
But since triangle $CED$ has angles of $60^o$, $30^o$, and $90^o$, then in triangle $FEG$
$$\angle FEG=2\angle CEG=60^o$$
And since
$$\angle HGE=\angle BGD=30^o$$
it follows that in triangle $HGE$ the angle at $H$ is right.
Hence $GH$ is the perpendicular bisector of $EF$, so that
$$\angle EFG=\angle FEG=60^o$$
and consequently triangle $EFG$ is equilateral.
But in an equilateral triangle, altitude, median, and angle bisector drawn from any vertex coincide, and those from all three vertices are concurrent at the circumcenter.
Hence equilateral triangle $ABC$ must be infinitely small compared to triangle $EFG$, i.e. a point. Therefore, a solution in the case where the three lines are drawn from distinct vertices seems impossible. And since an altitude, angle bisector, and median drawn from one vertex cannot intersect to form a triangle, any solution (see @MvG) must have two of the lines drawn from one vertex and the third from another.
A: Let's broaden the question slightly, and ask when we can create a triangle $\triangle PQR$ with (not-necessarily-equal) angle measures $p$, $q$, $r$.
We can situate $P$, $Q$, $R$ on the unit circle by taking coordinates
$$P := (1,0) \qquad Q := (\cos 2r, \sin 2r) \qquad R := (\cos 2q,-\sin2 q) \tag{1}$$

Consider first the case where the median, altitude, and bisector pass through separate vertices. Define the vertices of $\triangle ABC$ by
$$A = \frac{Q + a R}{1+a} \qquad B := \frac{R + b P}{1+b} \qquad C := \frac{P+c R}{1+c} \tag{2}$$
(where $a$, $b$, $c$ are dimensionless parameters, no side-lengths). Note that none of $a$, $b$, $c$ can be $-1$, as this puts the corresponding point "at infinity". (On the other hand, infinity is allowed as a parameter; if, for instance $a=\pm\infty$, then $A=R$.) Define
$$D := \overleftrightarrow{QR}\cap\overleftrightarrow{BC} \qquad
E := \overleftrightarrow{RP}\cap\overleftrightarrow{CA} \qquad
F := \overleftrightarrow{PQ}\cap\overleftrightarrow{AB} \tag{3}$$
We'll suppose that the median, altitude, and bisector pass through $A$, $B$, $C$, respectively. Thus, we require
$$D = \frac12(B+C) \qquad (E-B)\cdot(C-A) = 0 \qquad \frac{|AF|}{|BF|}=\frac{|AC|}{|BC|} \tag{4}$$
With the help of Mathematica, I find that conditions $(4)$ give

$$a = -\frac{\cos(p-q) \cos p \sin r}{\sin q\cos(2 p - r)} 
\qquad b = -\frac{\cos(p-q) \sin p}{2\cos p\sin r} \qquad
c = \frac{2 \cos 2 p \sin q}{\sin p \cos(p-q)} \tag{5}$$

When $p=q=r=60^\circ$, we get $a=b=c=-1$, making $\triangle ABC$ "infinitely large". (This is consistent with @Edward's answer, which finds that a finite $\triangle ABC$ leads to an "infinitely small" $\triangle PQR$.) Thus, we get no equilateral triangles. $\square$
More generally, we get $a=b=c=-1$ whenever 
$$\tan q = -\frac{\sin 2p}{1+3\cos 2p}$$ 
Also, $r = 90^\circ$ implies $a= -1$, so such right triangles are impossible, too. Everything else seems to be fair game. 
For fun, here are the six configuarations whre the angles of $\triangle PQR$ are $50^\circ$, $60^\circ$, $70^\circ$, in some order:
 
 
 

I have not yet completed analysis of the case where two of the segments pass through a common vertex. I will update.
