One strategy, probably not optimal, is to find the lengths of the sides, and then use Heron's Formula.
Take a triangle $XYZ$, with sides $x$, $y$, $z$ as usual, and let $m$ be the length of the median from $Z$ to the side $XY$, which has length $z$.
Let $P$ be the midpoint of $XY$. We have divided our triangle into two triangles, $ZPX$ and $ZPY$. Let $\theta=\angle ZPX$ and $\phi=\angle ZPY$. Note that $\theta+\phi=\pi$. By the Cosine Law, we have
$$y^2=m^2+\left(\frac{z}{2}\right)^2-mz\cos\theta.$$
Also by the Cosine Law, we have
$$x^2=m^2+\left(\frac{z}{2}\right)^2-mz\cos\phi.$$
But $\cos\phi=-\cos\theta$. Add. The awkward cosine terms cancel, and we get
$$x^2+y^2=2m^2+2\left(\frac{z}{2}\right)^2,$$
which yields the result
$$m^2=\frac{1}{4}\left(2x^2+2y^2-z^2\right).$$
The rest is downhill. We go back to the triangle of the question. The three lengths of the medians give us three linear equations in the squares of the side lengths. If the sides of our triangle are $a$, $b$, and $c$, we get
$$4\cdot 13^2=2a^2+2b^2-c^2;\quad 4\cdot 14^2=2b^2+2c^2-a^2;\quad 4\cdot 15^2=2c^2+2a^2-b^2.$$
Solve. A simple way is to use the symmetry of the equations by first of all "adding" the above three equations to first find $a^2+b^2+c^2$.
Now that we have the sides, we can use Heron's Formula.
This is my short proof from 1963:
In the triangle ABC draw medians BE, and CF, meeting at point G.
Construct a line from A through G, such that it intersects BC at point D.
We are required to prove that D bisects BC, therefore AD is a median, hence medians are concurrent at G (the centroid).
Proof:
Produce AD to a point P below triangle ABC, such that AG = GP.
Construct lines BP and PC.
Since AF = FB, and AG = GP, FG is parallel to BP. (Euclid)
Similarly, since AE = EC, and AG = GP, GE is parallel to PC
Thus BPCG is a parallelogram.
Since diagonals of a parallelogram bisect one another (Euclid),
therefore BD = DC.
Thus AD is a median. QED
Corollary: GD = AD/3.
Proof:
Since AG = GP and GD = GP/2, AG = 2GD.
AD = (AG + GD) = (2GD + GD) = 3GD.
Hence GD = AD/3. QED
Best Answer
In a triangle with sides $a$, $b$, and $c$, the medians to the sides of length $a$ and $b$ are perpendicular if and only if $a^2+b^2=5c^2$. But $11^2+13^2\ne5\cdot12^2$, so something is wrong with the question. https://en.wikipedia.org/wiki/Median_(geometry)#Other_properties