Giải thích các bước giải:
a.Ta có :$\widehat{ANB}=\widehat{APC}(=90^o)\to\Delta APC\sim\Delta ANB(g.g)$
$\to\dfrac{AP}{AN}=\dfrac{AC}{AB}\to AN.AC=AP.AB$
b.Ta có : $\widehat{PHB}=\widehat{NHC},\widehat{HPB}=\widehat{HNC}(=90^o)$
$\to\Delta HPB\sim\Delta HNC(g.g)$
$\to\dfrac{HP}{HN}=\dfrac{HB}{HC}$
Mà $\widehat{PHN}=\widehat{BHC}\to\Delta BHC\sim\Delta PHN(c.g.c)$
c.Từ câu a
$\to\dfrac{S_{ANP}}{S_{ABC}}=(\dfrac{AP}{AC})^2$
$\to\dfrac{S_{ANP}}{S_{ABC}}=(\cos\widehat{PAC})^2$
$\to\dfrac{S_{ANP}}{S_{ABC}}=(\cos60^o)^2$
$\to\dfrac{S_{ANP}}{S_{ABC}}=\dfrac14$
d.Ta có : $MI\perp AB, MK\perp BN\to MI//CP, MK//CN$
$\to\dfrac{BK}{BN}=\dfrac{BM}{BC}=\dfrac{BI}{BP}\to IK//PN(1)$
Tương tự $EF//PN(2)$
Vì $ME\perp CP, MI\perp AB, CP\perp AB$, HM\perp BC
$\to \widehat{HMB}=\widehat{HKM}(=90^o)\to \Delta HKM\sim\Delta HMB(g.g)$
$\to\dfrac{HK}{HM}=\dfrac{HM}{HB}\to HM^2=HK.HB$
Tương tự $HM^2=HE.HC$
$\to HK.HB=HE.HC\to\dfrac{HK}{HE}=\dfrac{HC}{HB}$
Lại có $\Delta BHC\sim\Delta PHN$ (câu b)
$\to\dfrac{HC}{HB}=\dfrac{HN}{HP}\to\dfrac{HK}{HE}=\dfrac{HN}{HP}$
$\to\dfrac{HK}{HN}=\dfrac{HE}{HP}\to KE//PN(3)$
Từ $(1),(2),(3)\to I,K,E,F$ thẳng hàng
e.Ta có : $\widehat{ANH}=\widehat{AMC}(=90^o)\to\Delta AHN\sim\Delta ACM(g.g)$
$\to\dfrac{AN}{AH}=\dfrac{AM}{AC}\to\Delta AHC\sim\Delta ANM(c.g.c)$
$\to\widehat{AMN}=\widehat{ACH}$
Tương tự $\widehat{AMP}=\widehat{ABN}$
Vì $\Delta HPB\sim\Delta HNC(câu b)$
$\to\widehat{PBH}=\widehat{HCN}\to\widehat{AMP}=\widehat{AMN}$
$\to MA$ là phân giác $\widehat{NMP}$
