If two parallel lines are cut by a transversal ,prove that the bisectors of the interior angles on the same side of the tranversal intersect each other at right angles

The figure shown below shows two parallel lines AB and CD cutting by a transversal l;

X and Y are the points of intersection of l with AB and CD respectively. XP, XQ, YP and YQ are the angle bisectors of $\angle \mathrm{AXY}, \angle \mathrm{BXY}, \angle \mathrm{CYX} \mathrm{and} \angle \mathrm{DYX} \mathrm{respectively}.$
$$\begin{gathered} \mathrm{AB}\parallel \mathrm{CD} \mathrm{and} \mathrm{l} \mathrm{is} \mathrm{a} \mathrm{transversal}, \\ \mathrm{So}; \angle \mathrm{AXY} = \angle \mathrm{DYX} (\mathrm{Pair} \mathrm{of} \mathrm{alternate} \mathrm{angles}) \\ \Rightarrow \frac{1}{2}\angle \mathrm{AXY} = \frac{1}{2}\angle \mathrm{DYX} \\ \Rightarrow \angle 1 = \angle 4 \\ \Rightarrow \mathrm{PX}\parallel \mathrm{YQ} ...\left(\mathrm{i}\right) \{\mathrm{If} \mathrm{a} \mathrm{transversal} \mathrm{intersects} \mathrm{two} \mathrm{lines} \mathrm{in} \mathrm{such} \mathrm{a} \mathrm{way} \mathrm{that} \mathrm{a} \mathrm{pair} \mathrm{of} \mathrm{alternate} \mathrm{interior} \mathrm{angles} \mathrm{are} \mathrm{equal}, \mathrm{then} \mathrm{two} \mathrm{lines} \mathrm{are} \mathrm{parallel}\} \\ \mathrm{Also}; \angle \mathrm{BXY} = \angle \mathrm{CYX} (\mathrm{Pair} \mathrm{of} \mathrm{alternate} \mathrm{angles}) \\ \Rightarrow \frac{1}{2}\angle \mathrm{BXY} = \frac{1}{2}\angle \mathrm{CYX} \\ \Rightarrow \angle 2 = \angle 3 \\ \Rightarrow \mathrm{PY}\parallel \mathrm{XQ} ,...\left(\mathrm{ii}\right) \{\mathrm{If} \mathrm{a} \mathrm{transversal} \mathrm{intersects} \mathrm{two} \mathrm{lines} \mathrm{in} \mathrm{such} \mathrm{a} \mathrm{way} \mathrm{that} \mathrm{a} \mathrm{pair} \mathrm{of} \mathrm{alternate} \mathrm{interior} \mathrm{angles} \mathrm{are} \mathrm{equal}, \mathrm{then} \mathrm{two} \mathrm{lines} \mathrm{are} \mathrm{parallel}\} \\ \mathrm{So} \mathrm{from} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \mathrm{we} \mathrm{get}; \\ \mathrm{PXQY} \mathrm{is} \mathrm{a} \mathrm{parallelogram}. ...\left(\mathrm{iii}\right) \\ \angle \mathrm{CYD} = 180° \\ \Rightarrow \frac{1}{2}\angle \mathrm{CYD} = 90° \\ \Rightarrow \frac{1}{2}\left(\angle \mathrm{CYX}+\angle \mathrm{DYX}\right) = 90° \\ \Rightarrow \frac{1}{2}\angle \mathrm{CYX}+\frac{1}{2}\angle \mathrm{DYX} = 90° \\ \Rightarrow \angle 3+\angle 4 = 90° \\ \Rightarrow \angle \mathrm{PYQ} = 90° \\ \mathrm{Hence} \mathrm{proved}. \end{gathered}$$