如图所示,▱ABCD中,AC与BD交于O点,E为AD延长线上一点,OE交CD于F,EO延长线交AB于G.求证:AB/DF−AD/DE=2.

问题描述:

如图所示,▱ABCD中,AC与BD交于O点,E为AD延长线上一点,OE交CD于F,EO延长线交AB于G.求证:

AB
DF
AD
DE
=2.

证明:延长CB与EG,其延长线交于H,如虚线所示,构造平行四边形AIHB.在△EIH中,由于DF∥IH,∴IHDF=EIED.∵IH=AB,∴ABDF=EIED,从而,ABDF-ADDE=EIED-ADED=EI−ADED=ED+AIED=1+AIED.①在△OED与△OBH中,∠DOE=...