相似三角形判定的题，求完整过程！！！

显示全部楼层 · 2013-1-20 13:38:27

过A作AG∥DC分别交EF、BC于G、H点，则四边形AGED、四边形GHCF是平行四边形
∴AG=DF=16／3，GH=FC=AE∵ EF∥BC ∴⊿AEG∽⊿ABH
∴AE／AB=AG／AH即AE／﹙AE＋BE﹚=AG／﹙AG＋GH﹚∴AE／﹙AE＋27／4﹚=﹙16／3﹚／﹙16／3＋AE﹚ 解得AE=±6 ﹙负数舍去﹚
∴AE=6...

千问 · 2013-1-20 13:38:27

作AG平行DC交BC于GEF于HAH=DFGH=FCAE=FCAB=AE+BEAG=AH+FC=DF+AEAH=AE/AB=AH/AGAE/(AE+BE)=DF/(DF+AE)...

千问 · 2013-1-20 13:38:27

作AG平行DC交BC于GEF于HAH=DFGH=FCAE=FCAB=AE+BEAG=AH+FC=DF+AEAE/AB=AH/AGAE/(AE+BE)=DF/(DF+AE)AE(DF+AE)=DF(AE+BE)AEDF+AEAE=DFAE+DFBEAEAE=DFBE=(6+3/4)(5+1/3)=36AE=6...