解：?$ (1)$?作法：以點(diǎn)?$B$?為圓心，?$AB$?長(zhǎng)半徑
作弧，交射線?$AC$?于點(diǎn)?$P，$?
?$P $?即為所求作的點(diǎn)
?$ (2)$?作法：?$①$?分別以點(diǎn)?$B、$??$A$?為圓心，
?$AB$?長(zhǎng)為半徑作弧交射線?$AC$?于點(diǎn)?$ Q_{1}、$??$Q_{2}$?
?$ ②$?作?$AB$?的垂直平分線，交射線?$AC$?于
點(diǎn)?$ Q_{3}$?
?$ ③$?連接?$ BQ_{1}、$??$BQ_{2}、$??$BQ_{3}$?
?$ △ABQ_{1}、$??$ △ABQ_{2}、$??$ △ABQ_{3}$?
即為所求作的三角形
