1. **Problem statement:** Given that lines $mn \parallel xy$ and angle $\widehat{ABy} = 60^\circ$, find angle $\widehat{BAm}$. Also, given that $At$ is the bisector of angle $nAB$ and $Bz$ is the bisector of angle $cBy$, prove that $At \parallel Bz$. 2. **Relevant formulas and rules:** - When two parallel lines are cut by a transversal, alternate interior angles are equal. - The sum of angles around a point on a straight line is $180^\circ$. - The angle bisector divides an angle into two equal parts. - If two lines are cut by a transversal and the alternate interior angles are equal, then the lines are parallel. 3. **Step-by-step solution:** **a. Calculate $\widehat{BAm}$:** - Since $mn \parallel xy$ and $AB$ is a transversal, angle $\widehat{ABy} = 60^\circ$ is given. - By the alternate interior angle theorem, $\widehat{BAm} = \widehat{ABy} = 60^\circ$. **b. Prove $At \parallel Bz$:** - $At$ is the bisector of angle $nAB$, so it divides $\widehat{nAB}$ into two equal angles. - $Bz$ is the bisector of angle $cBy$, so it divides $\widehat{cBy}$ into two equal angles. - Since $mn \parallel xy$, angles $\widehat{nAB}$ and $\widehat{cBy}$ are alternate interior angles and thus equal. - Therefore, the bisectors $At$ and $Bz$ create equal angles with the transversal $AB$. - Hence, $At \parallel Bz$ by the converse of the alternate interior angle theorem. **Final answers:** - $\boxed{\widehat{BAm} = 60^\circ}$ - $\boxed{At \parallel Bz}$