Solution: (1) problem, ∵AD//EF//BC

∴AE/EB=DO/BO

Equal proportion theorem (AE+EB)/EB=(D0+BO)/BO

∴AB/BE=BD/BO

And ∠ABD is a male horn.

∴△BEO∽△BAD

Executive Director

Similarly, △CFO∽△CDA (SAS)

∴FO/AD=CF/CD

∫AD//EF//BC

∴BE/AE=CF/DF

Equal ratio theorem BE/(AE+BE)=CF/(DF+CF)

∴BE/AB=CF/CD

∫FO/AD = CF/CD，EO/AD=BE/BA。

∴FO/AD=EO/AD

∴FO=EO

(2) ∵AD//EF//BC

∴AE/BE=AO/CO

Equal ratio theorem AE/(BE+AE)=AO/(CO+AO)

∴AE/AB=AO/AC

∵∠BAC is a male angle.

∴△AEO∽△ABC

∴EO/BC=AE/AB

EO/AD=BE/AB from ( 1)

∴eo/bc+eo/ad=ae/ab+be/ab= 1

(3) EO/BC+EO/AD=65438+ 0 from (2)

∴ 1/bc+ 1/ad= 1/oe

EO=FO is known from (1)

∴EO= 1/2EF， 1/EO=2/EF

∴ 1/BC+ 1/AD=2/EF