Question

如图,△ABC的内切圆⊙O切AC于点D,切BC于点F,切AB于点E,AD=4,角C=90°,⊙O的半径为2,求s△abc

Answer 1

∵⊙O为△ABC内切圆,r=2∴OE=OD=OF=2,OE⊥AB,OD⊥AC,OF⊥BC∴AE=AD=4,BE=BF,DC=CF (角分线)又∵角C=90°,可证四边形ODCF为正方形∴DC=CF=2设BE=EF=x则在RT△ABC中,（4+x）^2=6^2+(2+x)^2解得x=6所以S△ABC=1/2 * BC...