<BAR+<QCD=180 (opposite angles in a cyclic quad), let z=<QCD, <BAR=180-z;
Also, <ABQ+<RDC=180 (ditto), let y=<ABQ, <RDC=180-y.
Let x=<APR=<RPD (given because of bisected angle BPC).
<ARQ+<QRD=180 (angles on a straight line), let w=<ARQ, w+180-z+x=180 (triangle APR), so w=z-x.
<QCP=180-z (angles on straight line).
<BQR+<CQR=180 (ditto), let v=<BQR, <CQR=180-v=x+180-z (external angle of triangle PQC), so v=z-x. But w=z-x so v=w and <ARQ=<BQR (i) proved.
And since <BQR+<RQC=180, <ARQ+<RQC=180 (ii) proved.