解:由于O为△ABC内心,为角平分线的交点,则∠FAO=∠OAD,又圆O切于△ABC,D、E、F为切点,则∠OFA=∠ODA=90度,OF=OD,△AFO≌ADO,于是可设AF=AD=b,同理DB=BE设为a
同时又∠C=90°,则∠FCO=∠OCE=45°,则∠FOC=∠COE=45°
∴可设FO=OE=FC=CE=r
又∠C=90°,则依勾股定理有(b+r)^2+(a+r)^2=(a+b)^2
展开得b^2+r^2+2br+a^2+r^2+2ar=a^2+b^2+2ab
化简得2(r^2)+2br+2ar=2ab ,r^2+br+ar=ab
而r^2+br+ar=r(r+a+b)=(1/2)[(r+a)+(b+r)+(a+b)]r
=(1/2)[CB+AC+AB]r
=(1/2)rCB+(1/2)rAC+(1/2)rAB
=S△OCB+S△AOC+S△AOB
=S△ABC
又r^2+br+ar=ab等量代换,知=S△ABC=ab=AD×BD
命题得证.