One altitude, one bisector and one median are equal implies equilateral triangle Let $ABC$ be a triangle such that the lenghts of the $A$-angle-bisector, the $B$-median and the $C$-altitude are equal. Prove that $ABC$ is equilateral.
I was only able to show that $AB$ is the smallest side -- this follows from the $C$-altitude being the largest one. I cannot figure out a proper chain of inequalities to strengthen this argument, though I think it might be useful to use that from one vertex the angle bisector is always positioned between the median and the altitude.
Any help appreciated!
Update (thanks for the comments): Treat $ABC$ to be acute and if possible, try to give a synthetic proof (angle chasing, congruences, etc.)
 A: Let $x$ be the common $A$-bisector/$B$-median/$C$-altitude.
In its role as the $A$-bisector, $x$ satisfies (by Stewart's Theorem and the Angle Bisector Theorem)
$$b^2\,\frac{ac}{b+c}+c^2\,\frac{ab}{b+c}=a\left(x^2+\frac{a^2bc}{(b+c)^2}\right)\;\to\;x^2(b+c)^2=bc(a+b+c)(-a+b+c) \tag{1}$$
As the $B$-median (again by Stewart, or Apollonius),
$$c^2\;\frac{b}{2}+a^2\;\frac{b}{2} = b \left(x^2+\frac{b^2}{4}\right)\quad\to\quad
4x^2 = 2a^2-b^2+2c^2 \tag{2}$$
As the $C$-altitude (invoking Heron's formula),
$$\frac12cx=|\triangle ABC|\quad\to\quad4c^2x^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{3}$$
The non-linear system $(1)$, $(2)$, $(3)$ turns out to be quartic. Throwing it into Mathematica for numerical solution, and discarding non-positive values, gives two options:

$$(a,b,c,x) = \left(1,1,1,\frac12\sqrt{3}\right)\qquad (a,b,c,x) = \left(1,  1.2225\ldots, 0.2381\ldots, 0.3933\ldots\right) \tag{$\star$}$$

(The values in my comment to the question correspond to taking $c=1$.) The first solution is, of course, the equilateral triangle; the second is obtuse:

So, if we restrict ourselves to acute triangles, the equilateral becomes the only solution. $\square$
