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 $180^\circ$)?

This is the most basic approach, I believe. Consider the triangle $\triangle ABC$, in which $AD$ is both median and bisector.


*

*Extend $CD$ to a segment $DC'\cong CD$.

*Then $\triangle ACD \cong \triangle BDC'$ by SAS criterion.

*Consequently you have $\angle BC'D \cong \angle ACD$ and $AC \cong BC'$.

*For transitivity then $\angle BC'D \cong \angle BCD$.

*Thus $\triangle BCC'$ is isosceles and $BC \cong BC'$.

*But from transitivity and point 3. then $BC \cong AC$ and $\triangle ABC$ is isosceles.



