Question

如图,△ABC为等边三角形,AB=4,D是AC边上一动点,D不与A,C重合,EF垂直平分BD,设CD=x,AE=y.

Answer 1

(1)EF垂直平分BD

BE=ED

BF=FD

EF=EF

△BEF1≌△DEF

∠EDF=∠EBF=60

(2)由（1）知∠ADE+∠CDF=120=∠CDF+∠CFD

∠ADE=∠CFD

∠A=∠C

△AED∽△CDF

AE:AD=CD:CF

CF=4-BF=4-BE=4-(4-y)=y

y:(4-x)=x:y

y^2=4x-x^2

定义域0