50 nha 45p bắt đầu.

Giải thích các bước giải:

a.Xét $\Delta AHM,\Delta BOM$ có:

$\hat H=\hat O(=90^o)$

$\widehat{AMH}=\widehat{OMB}$

$\to \Delta AHM\sim\Delta BOM(g.g)$

$\to \dfrac{MA}{MB}=\dfrac{MH}{MO}$

$\to MA\cdot MO=MH\cdot MB$

b.Xét $\Delta AMB,\Delta HMO$ có:

$\widehat{AMB}=\widehat{HMO}$

$ \dfrac{MA}{MB}=\dfrac{MH}{MO}$

$\to\Delta MAB\sim\Delta MHO(c.g.c)$

$\to \widehat{AOH}=\widehat{MOH}=\widehat{MBA}=\widehat{ABC}=\widehat{ACB}$
c.Xét $\Delta SOA,\Delta SHB$ có:

chung $\hat S$

$\widehat{SOA}=\widehat{SHB}(=90^o)$

$\to \Delta SOA\sim\Delta SHB(g.g)$

$\to \dfrac{SO}{SH}=\dfrac{SA}{SB}$

Mà $\widehat{OSH}=\widehat{ASB}$

$\to \Delta SOH\sim\Delta SAB(c.g.c)$

$\to \widehat{SHO}=\widehat{SBA}$

Ta có: $\Delta ABC$ cân tại $A, AH\perp BC\to AH$ là trung trực $BC$

Mà $S\in AH$

$\to B, C$ đối xứng qua $AS$

$\to \widehat{ACS}=\widehat{ABS}=\widehat{OHS}=\widehat{AHK}$

Mà $\widehat{HAK}=\widehat{SAC}$

$\to \Delta AHK\sim\Delta ACS(g.g)$

$\to \dfrac{AH}{AC}=\dfrac{AK}{AS}$

Mà $\widehat{SAK}=\widehat{HAC}$

$\to \Delta AKS\sim\Delta AHC(c.g.c)$

$\to \widehat{AKS}=\widehat{AHC}=90^o$

$\to SK\perp AC$