The diagram shows a parallelogram ABCD. APQ, BRQ, CRS and DPS are straight lines which bisect angles A, B, C and D respectively.

1. Express the two required angles as halves of the parallelogram angles.

Since $AP$ bisects $\angle DAB$,
$\angle PAD=\frac12\angle DAB.$
Since $CR$ bisects $\angle BCD$,
$\angle RCB=\frac12\angle BCD.$

2. Use the property of opposite angles in a parallelogram to prove the equality.

In parallelogram $ABCD$, opposite angles are equal, so
$\angle DAB=\angle BCD.$
Therefore,
$\frac12\angle DAB=\frac12\angle BCD,$
so
$\angle PAD=\angle RCB.$

1. State the first pair of equal corresponding angles using the result from part (a).

From part (a),
$\angle PAD=\angle RCB.$

2. Prove the second pair of equal corresponding angles using angle bisectors and opposite angles of a parallelogram.

Since $DP$ bisects $\angle CDA$,
$\angle ADP=\dfrac12\angle CDA.$
Since $BR$ bisects $\angle ABC$,
$\angle CBR=\dfrac12\angle ABC.$
Opposite angles of a parallelogram are equal, so
$\angle CDA=\angle ABC.$
Therefore,
$\angle ADP=\angle CBR.$

3. Use the equal side of the parallelogram and conclude congruence by ASA.

In parallelogram $ABCD$, opposite sides are equal, so
$AD=CB.$
Thus, in $\triangle ADP$ and $\triangle CBR$,
$\angle PAD=\angle RCB,\qquad AD=CB,\qquad \angle ADP=\angle CBR.$
Therefore,
$\triangle ADP\cong \triangle CBR \quad \text{(ASA)}.$