Prove that <ARQ=<BQR

Question

 In the given figure, PR is the bisector of <BPC. It meets BC and AD at points Q and R respectively. Prove that:
i)<ARQ=<BQR

ii)<ARQ+<RQC=180º

Answer 1

<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.