Giải thích các bước giải:
a.Ta có : $BM\perp AC, CN\perp AB\to \widehat{AMB}=\widehat{ANC}=90^o$
$\to \Delta ABM\sim\Delta ACN(g.g)$
b.Ta có $\widehat{HNB}=\widehat{HMC}=90^o,\widehat{NHB}=\widehat{MHC}$
$\to \Delta HNB\sim\Delta HMC(g.g)$
$\to\dfrac{HN}{HM}=\dfrac{HB}{HC}\to HB.HM=HN.HC$
c.Gọi $S_{ABC}=S, S_{HAB}=S_3, S_{AHC}=S_2, S_{HBC}=S_1$
Ta có :
$\dfrac{HK}{AK}=\dfrac{S_{HBK}}{S_{ABK}}=\dfrac{S_{HKC}}{S_{ACK}}=\dfrac{S_{HBK}+S_{HKC}}{S_{ABK}+S_{AKC}}=\dfrac{S_{HBC}}{S_{ABC}}=\dfrac{S1}{S}$
Tương tự
$\to \dfrac{HM}{BM}=\dfrac{S_2}{S},\dfrac{HN}{CN}=\dfrac{S_3}{S}$
$\to \dfrac{HK}{AK}+\dfrac{HM}{BM}+\dfrac{HN}{CN}=\dfrac{S_1}{S}+\dfrac{S_2}{S}+\dfrac{S_3}S=1$