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 $180^\circ$)?
 A: 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.



A: 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.
A: From M, drop perpendiculars AP and AQ to AB and AC respectively. Then,
by AAS, $\triangle APM \cong \triangle AQM$.
[If AAS is not allowed, try reflect M about AB to H and reflect M about AC to K. Note that $\triangle AHM \cong \triangle AKM$.]

Go on to prove $\triangle BPM \cong \triangle CQM$ using RHS.
[If RHS is not allowed, try BPM and CQM are two equal circles. $\angle B$ will be equal to $\angle C$ because of equal chords on equal circles.] 
As a result, $\angle B = \angle C$.
A: Let $AD$ be a median and a bisector of $\Delta ABC$ and $E\in AD$ such that $D$ be a midpoint of $AE$.
Thus, since $\measuredangle ADC=\measuredangle BDE,$ we obtain:
$$\Delta ADC\cong \Delta EDB,$$ which gives
$$BE=CA$$ and since
$$\measuredangle BAD=\measuredangle CAD=\measuredangle BED,$$ we obtain $$AB=BE,$$ $$AB=AC$$ and we are done!
