(本题3分)(2022-浙江金华$cdot$九年级期中)如图,$DAperpAC,BCperpAC,AB$与$CD$相交于点$E$,过点$E$作$EFperpAC$交$AC$于$F$.且$BC=2,AD=3$,则$EF$的长为<ImageHere>
【分析】由$DAperpAC,BCperpAC,EFperpAC$,可得$EF//BC//AD$,则$triangleAEFsimtriangleABC,triangleCEFsimtriangleCDA$,再利用相似三角形的性质可得答案.【详解】解:$becauseDAperpAC,BCperpAC,EFperpAC$,$thereforeEF//BC//AD$,$thereforetriangleAEFsimtriangleABC,triangleCEFsimtriangleCDA$$thereforefrac{EF}{BC}=frac{AF}{AC},frac{EF}{AD}=frac{CF}{AC}$,$thereforefrac{EF}{BC}+frac{EF}{AD}=frac{AF}{AC}+frac{CF}{AC}=frac{AC}{AC}=1$,$becauseBC=2,quadAD=3$,$thereforefrac{EF}{2}+frac{EF}{3}=1$,解得:$EF=frac{6}{5}$.故答案为:$frac{6}{5}$.【点睛】本题考查的是相似三角形的判定与性质，证明$triangleAEFsimtriangleABC,triangleCEFsimtriangleCDA$是解本题的关键.