延长BD,作AE⊥BD,交其延长线于E,并与BC延长线交于F,
则〈FBE=〈EBA,(角平分线),
DE=DE,
〈AEB=〈FEB=90度,
△BFE≌△BAE,
〈F=〈EAB,
〈EDA=〈CDB(对顶角),
〈EAD=90度-〈EDA,
〈CBD=90度-〈CDB,
〈FAC=〈CBD=
RT△EAB≌RT△CFA,
BE=AC=12,
BD=8,
ED=12-8=4,
由前所述可知,
RT△EAB∽RT△CAF,
ED/CD=AD/BD,
AD*CD=ED*BD
设AD=x,
x(12-x)=4*8,
x^2-12x+32=0,
x1=8,x2=4,
取x1=8,(因CD不能为8),
CD=12-8=4,
BC^2=CD^2+BC^2,
BC=4√3,
∴S△ABC=AC*BC/2=12*4√3/2=24√3.