a ) Ta có : BC là đường kính của (O) \(\Rightarrow CM\perp AB,BN\perp AC\)
Mà \(BN\cap CM=H\Rightarrow H\) là trực tâm tam giác
\(\Rightarrow AK\perp BC\)
b ) Ta có : \(CM\perp AB,BN\perp AC\)
\(\Rightarrow\cos\widehat{A}=\frac{AM}{AC}=\frac{AN}{AB}\)
\(\Rightarrow AM.AB=AN.AC\)
c ) Ta có : \(AK\perp BC,BN\perp AC,CM\perp AB\)
\(\Rightarrow AMHN,MHKB,ANKB\) nội tiếp
\(\Rightarrow\widehat{KMH}=\widehat{KBH}=\widehat{KBN}=\widehat{KAN}=\widehat{HAN}=\widehat{HMN}\)
\(\Rightarrow MH\) là phân giác \(\widehat{NMK}\)
d ) Ta có :
\(\widehat{SMB}=\widehat{NCB}\left(+\widehat{BMN}=180^0\right)\)
\(\Rightarrow\Delta SMB\sim\Delta SCN\left(g.g\right)\Rightarrow\frac{SM}{SC}=\frac{SB}{SN}\Rightarrow SB.SC=SM.SN\)
Theo câu c ) \(\Rightarrow\widehat{NMK}=2\widehat{CMN}=2\widehat{NBC}=\widehat{NOC}\)
\(\Rightarrow MNOK\) nội tiếp
\(\Rightarrow\widehat{SKM}=\widehat{MNO}\)
\(\Rightarrow\Delta SMK\sim\Delta SON\left(g.g\right)\Rightarrow\frac{SM}{SO}=\frac{SK}{SN}\Rightarrow SM.SN=SK.SO\)
\(\Rightarrow SB.SC=SK.SO\)
