Question

如圖1-24,已知△ABC中,點(diǎn)D在CA延長線上,且,E為BC中點(diǎn),DE交AB于F,過點(diǎn)F引直線MN⊥DE,P為MN上一點(diǎn).?

求證:PD =PE.

圖1-24

Answer 1

思路分析:由于MN⊥DE,所以要證PD =PE,?只需證FD =FE,結(jié)合已知條件,過點(diǎn)E作EG∥AC交AB于點(diǎn)G,即可實(shí)現(xiàn).

證明:過點(diǎn)E作EG∥AC,交AB于點(diǎn)G.?

∵BE =EC,EG∥AC,?

∴BG =GA, .?

又∵,∴EG =DA.?

∵GE∥AC,∴∠ADF =∠GEF.

在△ADF和△GEF中,

∴△ADF≌△GEF.

∴DF =EF.?

又∵MN⊥DE,∴PD =PE.