(1)证明:S△ABC= AB·CD= BC·AE, ∵AB=2,BC=4, ∴ ×2×CD= ×4×AE, 即CD=2AE; (2)证明:如图②,连接PB, 则S△ABC=S△ABP+S△BCP, 即 AB·CD= AB·PF+ BC·PE, ∵AB=2,BC=4, ∴ ×2×CD= ×2×PF+ ×4×PE, 即CD=PF+2PE, 故2PE+PF=CD; (3)解:如图③,连接PB, 则S△ABP=S△ABC+S△PBC, 即 AB·PF= AB·CD+ BC·PE, ∵AB=2,BC=4, ∴ ×2×PF= ×2×CD+ ×4×PE, 即PF=CD+2PE。 | ![](http://img.shitiku.com.cn/uploads/allimg/20191024/20191024002656-15156.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191024/20191024002656-95483.png) |