By using midpoint theorem P is the midpoint of BC and CA is parallel to RC therefore I will also be the midpoint of DR and therefore we can say that DA equals to Dr

Given:
- P is the midpoint of side CD of parallelogram ABCD.
- Line through C parallel to PA intersects AB at Q and DA produced at R.

To Prove:
- DA = AR
- CQ = QR

Proof:

1. Midpoint Property:
- Consider the midpoint of a line segment. It divides the line segment into two equal parts.
- Since P is the midpoint of side CD, we have CP = PD.

2. Parallel Lines Property:
- When a line is drawn parallel to one side of a parallelogram, it intersects the other two sides in two distinct points.
- Hence, line CR intersects sides AB and AD produced.

3. Alternate Interior Angles Property:
- When a transversal intersects two parallel lines, the alternate interior angles are congruent.
- Therefore, ∠CQR = ∠PAC and ∠CQA = ∠PAR.

4. Corresponding Angles Property:
- When a transversal intersects two parallel lines, the corresponding angles are congruent.
- Therefore, ∠CQR = ∠PAC and ∠CQA = ∠PAR.

5. Congruent Triangles Property:
- From step 3 and 4, we can conclude that ∆CQR and ∆PAC are congruent by Angle-Side-Angle (ASA) congruence criteria.
- Since the corresponding sides of congruent triangles are equal, CQ = PA and QR = AP.

6. Parallelogram Property:
- In a parallelogram, opposite sides are equal in length.
- Therefore, CQ = AB and QR = AD.

7. Transitive Property:
- From step 5 and 6, we have CQ = AB = QR and PA = CQ = QR.

8. Congruent Triangles Property:
- Since P is the midpoint of CD, we have CP = PD.
- From step 2, we know that CR is parallel to PA.
- Therefore, ∆CAR and ∆DPA are congruent by Side-Angle-Side (SAS) congruence criteria.
- Hence, DA = AR.

Conclusion:
- From step 7, we have CQ = QR and PA = QR.
- From step 8, we have DA = AR.
- Therefore, DA = AR and CQ = QR.