If the median AM of a triangle ABC bisects the angle $\hat{A}$, then the triangle is an isosceles. Can we solve the above problem using only the criteria for congruent triangles (i.e., without using the fact that the sum of the angles of a triangle is $180^\circ$)?

Yes. If we have a $\triangle ABC$ with a point $M$ on $BC$, and $\angle A$ is bisected by $AM$, then we obtain $\angle BAM \cong \angle CAM$ and the 2 new triangles $\triangle BAM$ and $\triangle CAM$. Now, draw a line segment $PQ$ through $A$ parallel to $BC$. Then, draw line segments $PB$ and $QC$ perpendicular to $PQ$. 
Now, due to the alternate interior angles theorem, $\angle BAM \cong \angle PBA$ and $\angle CAM \cong \angle ACQ$. Now, $\angle ACQ + \angle ACM = 90$ and $\angle PBA + \angle BAM = 90$. 
I think from here the pieces should fit. Though if you have issues drawing the picture, I can provide one. Also, I wasn't sure whether you could use transversals or not, but considering how most geometry textbooks are, I assumed that perhaps you could.
