?$?解：(1)在Rt△ABC中，由勾股定理得：?$?
?$?AB2=BC2+AC2?$?
?$?即AB=\sqrt{AC2+BD2}=\sqrt{52+122}=13?$?
?$?(2)S_{△ABC}=\frac {1}{2}AC·BC=\frac {1}{2}AB·CO?$?
?$?即CO=\frac {AC·BC}{AB}=\frac {60}{13}?$?
?$?在Rt△ACO中，由勾股定理得：?$?
?$?AC2=AO2+CO2?$?
?$?即AO=\sqrt{AC2－CO2}=\sqrt{52－(\frac {60}{13})2}=\frac {25}{13}?$?
