Question

如图,在△ABC中,点D、E分别在AB、AC上,DE//BC,点F在BC边上,且∠BAF=∠DEF,AF与DE相交于点G

Answer 1

证明：AF交DE于O,

∠BAF=∠DEF,∠AOD=∠EOF.△AOD∽△EOF

AD：EF=AO：OE

OE‖CF,容易得△AOE∽△AFC,AF：CF=AO：OE

所以AD：EF=AF：CF,即AD×CF=EF×AF