1. **Problem statement:** Given trapezium WXYZ with WX \parallel ZY, rhombus AXYB inside it, and lines XBC and XAZ straight. Given angles: $\angle XAB=82^\circ$, $\angle ZYB=19^\circ$, $\angle WZX=57^\circ$, and $BC=BY$. Find $\angle c$ and $\angle w$. 2. **Key properties and formulas:** - In a rhombus, all sides are equal and opposite angles are equal. - Since AXYB is a rhombus, $AX=XY=YB=BA$. - $BC=BY$ implies triangle $BXC$ is isosceles with $BC=BY$. - $WX \parallel ZY$ means alternate interior angles are equal. 3. **Find $\angle c$: (angle at point C)** - Since $XBC$ is a straight line and $BC=BY$, triangle $BXC$ is isosceles with $BC=BY$. - $\angle ZYB=19^\circ$ is given, and since $BY=BC$, $\angle BXC=\angle ZYB=19^\circ$. - $\angle c$ is the angle at $C$ adjacent to $BXC$, so $\angle c=19^\circ$. 4. **Find $\angle w$: (angle at point W)** - Given $\angle WZX=57^\circ$ and $WX \parallel ZY$, by alternate interior angles, $\angle WXZ=57^\circ$. - In trapezium $WXYZ$, sum of interior angles on the same side of the transversal $XZ$ is $180^\circ$. - So, $\angle W + \angle WZX = 180^\circ$. - Substitute $\angle WZX=57^\circ$, so $\angle w = 180^\circ - 57^\circ = 123^\circ$. **Final answers:** $$\angle c = 19^\circ$$ $$\angle w = 123^\circ$$