Here, ABCD is trapezium.

Hence, AB || DC

Also given that AP = PD and BQ = CQ

(i) In ∆QCD and ∆QBE, we have,
∠DQC = ​∠BQE …Vertically opposite angles

∠DCQ = ∠EBQ …Alternate angles with transversal BC
BQ = CQ … P is the midpoint

Hence, by AAS test of congruency,
∆QCD ≅ ∆QBE
​Hence, DQ = QE …by cpct

(ii) Also, in ∆ADE, P and Q are the midpoints of AD and DE respectively

∴ PQ || AE

Hence, PQ || AB || DC

ie. AB || PR || DC

(iii) PQ, AB and DC are cut by transversal AD at P such that AP = PD.
Also they are cut by transversal BC at Q such that BQ = QC.
​ Similarly, lines PQ, AB and DC are also cut by AC at R.

Hence, by intercept theorem,
∴ AR = RC