TITLE: Triangles congruency
QUESTION [8 upvotes]: If one angle of two triangles and the two medians to the two sides forming this angle are equal, will the two triangles be congruent? For example, in triangles $ABC$ and $A'B'C'$, we have angle $A = A'$ and medians $BD = B'D'$ and $CE = C'E'$.
I have tried with Geogebra and indeed they are congruent. I used the fact that the 3 medians intersect at a point that has a distance of $\frac{1}{3}$ of its length from the origin vertex. By keeping one median fixed, I rotated the second one around the intersecting point and indeed there are two points which "see" the side $BC$ by a given angle, both of them resulting to the same triangles.
How do I solve it geometrically? 
Let's name the intersection point of the 3 medians $O$. I started by assuming that $BC \neq B'C'$, then since the other two sides of the small triangles (formed by the $\frac{2}{3}$ of the medians) are equal, then angle $BOC \neq B'O'C'$. Then also angles $DOE \neq D'O'E'$, so, segment $DE \neq D'E'$. But I don't know how to go further. 
Many thanks!

REPLY [2 votes]: Consider the Figure above, where I used your notation, and at the beginning I followed a path similar to the one suggested by Michael in his answer.

Draw segment $BD$ (the first given median) and let $O$ be the centroid of the desired triangle, so that $\overline{BO} = 2\overline{OD}$.
Draw the unique (up to reflection with respect to $BD$) circumference $\gamma_1$ passing through $B$ and $D$ and such that the chord $BD$ subtends on it the desidered angle $\angle BAD$.
Draw then the circle $\gamma_2$ passing through the midpoint of $BD$, and internally tangent to $\gamma_1$ in $B$. This is the locus of the midpoints of any chord on $\gamma_1$ having one endpoint in $B$.
Draw $\gamma_3$, the circle centered in $O$ with radius equal to $\frac13\overline{CE}$, where $CE$ is the second given median. 
$\gamma_3$ has at most one intersection point with $\gamma_2$ (in the half-plane where $A$, too, must lie). 

Now it suffices to demonstrate that the above mentioned intersection point between $\gamma_3$ and $\gamma_2$ is in fact the uniquely determined point $E$.
In particular, since $BE \cong AE$ by our construction, we need to show that $C = EO \cap AD$ is such that $AD \cong DC$.
Consider triangle $\triangle ABD$ cut by $EC$. By Menelaus's Theorem you have 
$$\frac{\overline{BO}}{\overline{OD}}\cdot \frac{\overline{DC}}{\overline{AC}}\cdot \frac{\overline{AE}}{\overline{BE}} = 1,$$
hence
$$\overline{AC} = 2\overline{DC},$$
and the thesis follows. $\blacksquare$