Question

△ABC的内切圆圆o与AC.BC,AB交于点DFE AB=6 AC=7BC=11则AE=?BF=?CD=?

Answer 1

因为△ABC的内切圆圆o与AC.BC,AB交于点DFE

则

AE=AD BE=BF CD=CF

AE+BE=AB=6 1

BF+CF=BC=11 2

AD+CD=AC=7 3

1+2+3得

AE+BE+BF+CF+AD+CD=2AE+2BF+2CF=24

AE+BF+CF=12 4

4-1得

CF=6

4-2得

AE=1

4-3得

BF=4