解：?$(1) \angle B P C=\angle A B P+\angle A C P+\angle B A C $?
理由: 如圖 ①, 連接?$ A P $?并延長至點(diǎn)?$ D . $?
∵?$\angle B P D= \angle A B P+\angle B A P，$??$ \angle C P D=\angle A C P+\angle C A P ，$?
?$\angle B A C=\angle B A P+\angle C A P，$??$ \angle B P C=\angle B P D+ \angle C P D，$?
∴?$\angle B P C=\angle A B P+\angle A C P+\angle B A C .$?
?$(2)$?如圖 ②, 連接?$ B C ，$? 連接?$ A P $?并延長至點(diǎn)?$ Q . $?
∵?$B A 、$??$ C A $?分別平分?$ \angle D B P 、$??$ \angle E C P $?交于點(diǎn)?$ A ，$?
∴?$\angle D B A=\angle A B P，$??$ \angle E C A=\angle A C P . $?
∵?$m / / n ，$? ∴?$\angle D B A+\angle A B P+\angle P B C+\angle E C A+\angle A C P+ \angle P C B=180° . $?
∵?$\angle B P C+\angle P B C+\angle P C B=180° ，$?
∴?$\angle B P C=\angle D B A+\angle A B P+\angle E C A+\angle A C P= 2 \angle A B P+2 \angle A C P . $?
∴?$\angle A B P+\angle A C P=\frac {1}{2} \angle B P C= 40° . $?
由?$ (1) $?可知?$ \angle B P C=\angle A B P+\angle A C P+\angle B A C ，$?
∴?$\angle B A C=\angle B P C-\angle A B P-\angle A C P=80°- 40°=40° $?
?$(3)$?如圖③, 連接?$ B C . $?
∵?$B F_1 、$??$ C F_1 $?分別平分?$ \angle G B P 、$??$ \angle H C P $?交于點(diǎn)?$ F_1，$?
∴?$\angle G B F_1=\angle P B F_1 ，$??$ \angle H C F_1=\angle P C F_1 .$?
∵?$m / / n，$? ∴?$\angle G B F_1+\angle F_1\ \mathrm {B} C+ \angle B C F_1+\angle H C F_1=180° . $?
∵ 在?$ \triangle B C F_1 $?中，?$ \angle F_1\ \mathrm {B} C+ \angle B C F_1+\angle F_1=180°，$?
∴?$\angle G B F_1+\angle H C F_1= \angle F_1 . $?
∵ 在?$ \triangle P B C $?中，?$ \angle P+\angle P B C+\angle P C B= 180°，$?
∴?$\angle F_1\ \mathrm {B} C+\angle B C F_1+\angle F_1+\angle P+\angle P B C+ \angle P C B=360° . $?
∴?$\angle P B F_1+\angle P C F_1+\angle F_1+\angle P= 360° . $?∴?$\angle F_1+\angle P+\angle F_1=360° .$?
∵?$\angle P=80° ，$? ∴?$\angle F_1=\frac {360°-80°}{2}=140° . $?
利用?$ (2) $?的結(jié)論：?$ \angle F_2= \frac {1}{2} ×140°=70° ，$? 以此類推，?$ \angle F_{2020}=\frac {140°}{2^{2019}}$?
