解:過?$C$?作?$CE⊥AB$?于?$E$?
∵?$CE⊥AB,$?∴?$AD=2\ \mathrm {A}E$?
∵?$∠C=90°,$??$AC=9,$??$BC=12$?
∴?$AB= \sqrt {AC^2+BC^2}=15$?
∵?$S_{△ABC}=\frac 12\ \mathrm {A}C· BC = \frac 12\ \mathrm {A}B· CE$?
∴?$CE= \frac {AC×BC}{AB}= \frac {9×12}{15}= \frac {36}5$?
在?$Rt?AEC$?中,?$AE= \sqrt {AC^2-CE^2}= \frac {27}5 $?
∴?$AD= 2\ \mathrm {A}E= \frac {54}5$?