解：?$(2)2∠F+∠P=180° $?
理由：∵?$ AF{平分}∠BAP，CF{平分}∠DCE，$?∴?$ ∠BAF=\frac {1}{2}∠BAP，$??$∠DCF=\frac {1}{2}∠DCE. $?
∵?$ AB//CD，$?∴?$ ∠BAF=∠DQF. $?
∵?$ ∠DQF$?是?$△CFQ$?的外角，∴?$ ∠F=∠DQF-∠DCF=∠BAF-∠DCF=\frac {1}{2} ∠BAP - \frac {1}{2}∠DCE$?
?$=\frac {1}{2} (∠BAP_{-}∠DCE)=\frac {1}{2} [∠BAP-(180°-∠DCP)]=\frac {1}{2} (∠BAP+∠DCP-180°).$?
由?$(1)$?可得，?$∠P+∠BAP+∠DCP=360°，$?∴?$ ∠BAP+∠DCP=360°-∠P. $?
∴?$ ∠F=\frac {1}{2} (360°-∠P-180°)=90°- \frac {1}{2}∠P，$?即?$2∠F+∠P=180°.$?
