专家们请帮忙解初二证明题我10-
知锐角三角形ABC,P为BC上一点,证明:AB^·PC+AC^·PB=BC·(AP^+PB·PC）
证明:作AD⊥BC于D(如图)
AB^=AD^+BD^
∴AB^·PC=(AD^+BD^)·PC……①
AC^=AD^+CD^
∴AC^·PB=(AD^+CD^)·PB……②
①+②得:
AB^·PC+AC^·PB=(AD^+BD^)·PC+(AD^+CD^)·PB
=AD^·PC+AD^·PB+BD^·PC+CD^·PB
=AD^·(PC+PB)+BD^·(BC-PB)+CD^·PB
=AD^·BC+BD^·BC+CD^·PB-BD^·PB
=AD^·BC+BD^·BC+(CD^-BD^)...全部
知锐角三角形ABC,P为BC上一点,证明:AB^·PC+AC^·PB=BC·(AP^+PB·PC）
证明:作AD⊥BC于D(如图)
AB^=AD^+BD^
∴AB^·PC=(AD^+BD^)·PC……①
AC^=AD^+CD^
∴AC^·PB=(AD^+CD^)·PB……②
①+②得:
AB^·PC+AC^·PB=(AD^+BD^)·PC+(AD^+CD^)·PB
=AD^·PC+AD^·PB+BD^·PC+CD^·PB
=AD^·(PC+PB)+BD^·(BC-PB)+CD^·PB
=AD^·BC+BD^·BC+CD^·PB-BD^·PB
=AD^·BC+BD^·BC+(CD^-BD^)·PB
=AD^·BC+BD^·BC+(CD+BD)·(CD-BD)·PB
=AD^·BC+BD^·BC+BC·(CD-BD)·PB
=BC·[AD^+BD^+(CD-BD)·PB]
=BC·[(AP^-PD^)+(PB+PD)^+(PC-PD-PB-PD)·PB]
=BC·[AP^-PD^+PB^+2PB·PD+PD^+PC·PB-PD·PB-PB^-PD·PB]
=BC·(AP^+PB·PC）
。
收起
