Giải thích các bước giải:
a.Xét $\Delta ABN,\Delta ACP$ có:
Chung $\hat A$
$\widehat{ANB}=\widehat{APC}=90^o$
$\to\Delta ABN\sim\Delta ACP(g.g)$
$\to \dfrac{AB}{AC}=\dfrac{AN}{AP}$
$\to \dfrac{AB}{AN}=\dfrac{AC}{AP}$
Mà $\widehat{PAN}=\widehat{BAC}$
$\to\Delta ANP\sim\Delta ABC(c.g.c)$
$\to \dfrac{NP}{BC}=\dfrac{AN}{AB}$
b.Xét $\Delta AHP, \Delta AMB$ có:
Chung $\hat A$
$\widehat{APH}=\widehat{AMB}(=90^o)$
$\to\Delta APH\sim\Delta AMB(g.g)$
$\to \dfrac{AP}{AM}=\dfrac{AH}{AB}$
$\to AP.AB=AH.AM$
Xét $\Delta AHB,\Delta APM$ có:
Chung $\hat A$
$AH.AM=AP.AB\to \dfrac{AP}{AH}=\dfrac{AM}{AB}$
$\to \Delta ABH\sim\Delta AMP(c.g.c)$
$\to \widehat{AHB}=\widehat{AMP}$
c.Ta có $\Delta ANP\sim\Delta ABC(cmt)$
$\to \dfrac{S_{ANP}}{S_{ABC}}=(\dfrac{AN}{AB})^2$
Mà $\widehat{ANB}=90^o,\widehat{NAB}=\hat A=60^o$
$\to \Delta ABN$ là nửa tam giác đều
$\to AN=\dfrac12AB\to \dfrac{AN}{AB}=\dfrac12$
$\to \dfrac{S_{ANP}}{S_{ABC}}=\dfrac14$
d.Ta có $NF//AP$
$\to \dfrac{PF}{FC}=\dfrac{NA}{NC}$
Gọi $PE\cap BC=D$
Vì $PE//AC$
$\to \dfrac{PE}{AN}=\dfrac{BE}{BN}=\dfrac{ED}{CN}$
$\to \dfrac{EP}{ED}=\dfrac{NA}{NC}$
$\to \dfrac{EP}{ED}=\dfrac{FP}{FC}$
$\to EF//CD\to EF//BC$
