?$(1)$?證明：∵?$∠A+∠B+∠AOB = 180°，$??$∠C+∠D+∠COD = 180°$?
∴?$∠A+∠B+∠AOB=∠C+∠D+∠COD$?
∵?$∠AOB=∠COD，$?∴?$∠A+∠B=∠C+∠D$?
解：?$(2)$?∵?$AP，$??$CP $?分別平分?$∠BAD，$??$∠BCD$?
∴?$∠BAP=∠P AD，$??$∠BCP=∠P CD$?
?$ $?由?$(1)$?的結(jié)論，得?$∠P+∠BCP=∠ABC+∠BAP ①，$?
?$∠P+∠P AD=∠ADC+∠P CD ②$?
①+②，得?$2∠P+∠BCP+∠P AD=∠BAP+∠P CD+∠ABC+∠ADC$?
∴?$2∠P=∠ABC+∠ADC$?
∵?$∠ABC = 36°，$??$∠ADC = 16°，$?∴?$2∠P=36°+16°=52°$?
∴?$∠P = 26°$?
?$ (3) ∠P = 90°+\frac 12(∠B+∠D)，$?理由如下：
∵直線(xiàn)?$AP $?平分?$∠BAD，$??$CP $?平分?$∠BCD$?的外角?$∠BCE$?
∴?$∠P AB=∠P AD，$??$∠P CB=∠P CE$?
∴?$2∠P AB+∠B = 180°-2∠P CB+∠D$?
∴?$180°-2(∠P AB+∠P CB)+∠D=∠B$?
∵?$∠P+∠P AD=∠P CB+∠AOC=∠P CB+∠B + 2∠P AD$?
∴?$∠P=∠P AD+∠B+∠P CB=∠P AB+∠B+∠P CB$?
∴?$∠P AB+∠P CB=∠P-∠B$?
∴?$180°-2(∠P-∠B)+∠D=∠B，$?即?$∠P = 90°+\frac 12(∠B+∠D)$?
