Euclidean triangle is determined by Angle, Median and radius of exterior circle consider two triangles $\Delta(A,B,C), \Delta(A',B',C')$ in euclidean plane. I want to prove that these triangles are congruent if they are equal in the following data:


*

*they have the same angle at one vertex, i.e. $\angle (BAC)\cong\angle (B'A'C')$

*the medians $s_A,s_{A'}$ joining $A, A'$ to the midpoint of the opposite sides $CB, C'B'$ are equal, i.e. $s_A\cong s_{A'}$

*the radius $R$ of the excircles, tangent to the sides $BC, B'C'$, are the same.
Now I want to prove that the remaining angles and all corresponding edges of the triangles must be congruent to each other. Do you have an idea how one can prove this? Unfortunately I couldn't prove the claim yet and need some help. 
Any help will be very much appreciated!
Best wishes
 A: Unfortunately, we can construct a counterexample,
where this information 
(the angle $\angle BAC$, 
the length of the median $|AM|$
and the radius $X_a$ of the excircle)
is not enough to define a unique triangle.
Let's start with an excircle $\mathcal{X}$
centered at the origin with the radius $r_a=3$,
and fix the point $A=(-5,0)$:

It can be shown that the locus of the midpoints of sides $BC$
where $B\in AP$, $C\in AQ$ is a hyperbola $\mathcal{H}$,
that passes through the points $D$ and $E$.
Note that $AD=AE=2$, 
that is $\mathcal{H}$ intersects 
a circle $\mathcal{A}$  around $A$ with the radius $2$ 
in two distinct points. 
Observe that $\mathcal{H}$ and $\mathcal{A}$
are very close, they are almost coincide with each other,
so while $\mathcal{H}$ splits any tangent $B_iC_i$ 
exactly at the midpoint, 
$\mathcal{A}$ 
will split the same tangent $B_iC_i$
very close to the midpoint, 
which can be addressed to the rounding error. 
However, if we slightly decrease 
the radius of $\mathcal{A}$
by amount $d_r$ in a small range, 
we can get many pairs 
of exact intersection points $M=\mathcal{X}\cap\mathcal{A}$ 
and $M'=\mathcal{X}\cap\mathcal{A}$
and construct a pair of very different triangles 
$\triangle ABC$ and $\triangle AB'C'$,
which have the same $\angle BAC=\angle B'AC'$,
the same excircle $\mathcal{X}$
and the same sized medians $|AM|=|AM'|=2-d_r$,
but the other angles and sides are very different 
(an example pair of $M,M'$ 
corresponding to $d_r=\frac{3}{256}$
is shown in the picture).
