In the diagram below, we have AB || DC, PQ = QR, BPQ = 100 + x, and APR = x, Find x in degress

Since AB || CD, angle(APR) = angle(QRP)              [alternate interior angles]

Since PQ = QR, angle(QRP) = angle(QPR)             [base angles of an isosceles triangle]

Since angle(APR) = x, angle(QRP) = x                    [substitution, or transitivity]

angle(BPQ) + angle(QPR) + angle(RPA) = 180°      [they form a line]

Therefore:  (100° + x) + (x) + (x) = 180°

                               100° + 3x  =  180°

                                           3x  =  80°

                                             x  =  (80/3)°