證明：?$(1)$?∵?$PA$?與?$\odot O$?相切于點?$A，$?
∴?$PA\perp OA.$?
∵?$PO$?平分?$∠APD，$??$OB\perp PD，$?
∴?$OB = OA，$?
∴?$PB$?是?$\odot O$?的切線?$.$?
?$(2)$?根據題意，得?$OA = OB = 4$?
.∵?$OC = 5，$?
∴?$AC = OA + OC = 4 + 5 = 9.$?
∵?$PA\perp OA，$??$OB\perp PD，$?
∴?$∠PAO=∠PBO=∠OBC = 90°，$?
∴在?$Rt\triangle OBC$?中，?$BC=\sqrt {OC^2-OB^2} = 3.$?
在?$Rt\triangle PAO$?和?$Rt\triangle PBO$?中，
?$OA = OB，$?
?$OP = OP，$?
∴?$Rt\triangle PAO\cong Rt\triangle PBO，$?
∴?$PA = PB.$?
∵在?$Rt\triangle PAC$?中，?$PA^2+AC^2=PC^2，$?
∴?$PA^2+9^2=(PA + 3)^2，$?
?$ \begin {aligned}PA^2+81&=PA^2+6PA + 9\\6PA&=72\\PA&=12\end {aligned}$?
∴?$PA$?的長是?$12.$?
