Circle Geometry Problem (1 Viewer)

[qqmore]

Please help me with this problem =(.......... harder 3u isnt my forte.... p.s. it is attached.

 

[lolokay]

< CSP = < RSQ (vert. opp)
< SCP = < RAP (exterior of opp side of cyclic quad)
< RPA = < DPR (given)
.'. RPA ||| SCP
with < ARP = < CSP = <QSR = <QRS
< QRY = QSY (given)
so < QYR = < QYS = 90' (angle sum of triangle, QYR+QYS = 180)
.'. QY is perpendicular to RP

 

[qqmore]

Cheers

 

[bored of sc]

(i)
< APR = < DPR --> given PR bisects angle APD
< DAP = < SCP --> exterior angle of cyclic quad ABCD = opposite interior angle
Therefore triangle SCP is similar to triangle RAP --> equiangular
< QRS = < CSP --> equal angles in similar triangles SCP and RAP
But < QSP = < CSP --> vertically opposite angles
Therefore < QRS = < QSP