Given: $▱\mathrm{JQKS};\overline{PJ}\cong \overline{RK}$, Prove: $\angle \mathrm{P}\cong \angle \mathrm{R}$.

The opposite angles of a parallelogram are congruent.

In $▱JQKS$, angles KSJ and KQJ are opposite angles. Therefore, $\angle KSJ\cong \angle KQJ$.

From the given figure, it can be observed that $\overline{SK}\parallel \overline{JQ}$and $\overline{KQ}$is a transversal then $\angle 2$and $\angle KQJ$are alternate interior angles such that, $\angle 2\cong \angle KQJ$.

From the given figure, it can be observed that $\overline{SK}\parallel \overline{JQ}$and $\overline{SJ}$is a transversal then $\angle 1$and $\angle KSJ$are alternate interior angles such that, $\angle 1\cong \angle KSJ$.

As $\angle 1\cong \angle KSJ$, $\angle 2\cong \angle KQJ$and $\angle KSJ\cong \angle KQJ$it implies that $\angle 1\cong \angle 2$.

The opposite sides of a parallelogram are congruent.

In $▱JQKS$, sides $\overline{SJ}$and $\overline{KQ}$are opposite sides. Therefore, $\overline{SJ}\cong \overline{KQ}$.

As $\overline{PJ}\cong \overline{RK}$, $\angle 1\cong \angle 2$and $\overline{SJ}\cong \overline{KQ}$then by SAS postulate $\Delta SPJ\cong \Delta KRQ$.

As $\Delta SPJ\cong \Delta KRQ$then by corresponding parts of congruent triangles, $\angle P\cong \angle R$.

Hence it is proved that $\angle P\cong \angle R$.