In the diagram below, we have \(\overline{AB}\parallel\overline{CD}\), \(QP = QR\), \(\angle BPQ = x^\circ + 100^\circ\), and \(\angle APR = x^\circ\). Find x.
Since QP = QR, 2x + 100 = 180. Then x = 40.
BPQ + QPR + APR = 180
so x+100+QPR+x = 180
or 2x + QPR = 80 ...(1)
QRP = APR
QPR = QRP (because triangle QRP is isosceles with QR = QP) hence QRP = APR = x
use this in (1) to get:
2x + x = 80
3x = 80
x = 80/3
