解：?$(2)AC+AD$?為定值，理由如下：
由題意得?$AB//OP//O'P'$?
∵?$AB//OP$?
∴?$△ABC∽△OPC$?
∴?$\frac {AB}{OP}=\frac {AC}{OC}$?
∵?$AB=h，$??$OP=O'P'=l，$??$OA=a$?
∴?$\frac h{l}=\frac {AC}{a+AC}$?
∴?$AC=\frac {ah}{l-h}$?
同理可得?$AD=\frac {(m-a)h}{l-h}$?
∴?$AC+AD=\frac {mh}{l-h}$?
∴?$AC+AD$?為定值
?$(3)$?設(shè)點(diǎn)?$A$?到點(diǎn)?$O$?的距離為?$S_{1}，$?點(diǎn)?$A$?到影子頂端?$C$?的距離為?$S_{2}$?
∵?$AB//OP$?
∴?$△ABC∽△OPC$?
∴?$\frac {AB}{OP}=\frac {AC}{OC}$?
∵?$AB=h，$??$OP=l，$??$AC=S_{2}，$??$OC=OA+AC=S_{1}+S_{2}$?
∴?$\frac h{l}=\frac {S_{2}}{S_{1}+S_{2}}$?
∴?$\frac l{h}-1=\frac {S_{1}}{S_{2}}$?
∴?$\frac {S_{1}}{S_{2}}=\frac {v_{1}}{v_{2}}=\frac {l-h}h$?
∴?$v_{2}=\frac {hv_{1}}{l-h}$?
