初二的一道几何题,

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初二的一道几何题,

1图连接AP,设AB=AC=a,则S⊿ABC = S⊿ABP+S⊿ACP
a*CF = a*PD+a*PE -> CF = PD+PE
2图连接AP,则S⊿ABC = S⊿ABP - S⊿ACP
a*CF = a*PD-a*PE -> CF = PD-PE
3图 连接AP,设AB=AC=a,则S⊿ABC = S⊿ABP+S⊿ACP
a*CF = a*PD+a*PE -> CF = PD+PE
4图连接AP,则S⊿ABC = S⊿ABP - S⊿ACP
a*CF = a*PD-a*PE -> CF = PD-PE
5图 连接AP,设AB=AC=a,则S⊿ABC = S⊿ABP+S⊿ACP
a*CF = a*PD+a*PE -> CF = PD+PE
6图连接AP,则S⊿ABC = S⊿ABP - S⊿ACP
a*CF = a*PD-a*PE -> CF = PD-PE
所以,P点在三角形内,CF = PD+PE;P点在三角形外,CF = PD-PE;