Two circles with centres P and Q intersect at B and C. A, D are points on the circles with centres P and Q respectively such that A, C, D are collinear. If LAPB = 130°, and LBQD = x, then the value of x is

Let angle PAB= z
=> angle QAB=180-z
Join QB to a point R on right circle such that A and R are in opposite segments. Now, AQRB is a cyclic quadrilateral.
=> angle QAB +angle QRB=180
=>angle QRB = z
Now using the property that angle subtended by an arc on circle is half that suspended at centre, we get angle BDQ =2z which gives angle BQD = 90-z.
Following similar procedure on left circle we get angle PCB=2z and angle
BPC= 90-z.
Hence we get angle BQD=angle BPC=90-z