Giải thích các bước giải:

a.Vì $SB, SC$ là tiếp tuyến của $(O)$

$\to SO\perp BC=M$

$\to SM.SO=SB^2$

Mà $\widehat{SBP}=\widehat{BAP}=\widehat{BAS}$

$\to \Delta SBP\sim\Delta SAB(g.g)$

$\to \dfrac{SB}{SA}=\dfrac{SP}{SB}$

$\to SP.SA=SB^2=SM.SO$

$\to \dfrac{SP}{SM}=\dfrac{SO}{SA}$

$\to \Delta SPM\sim\Delta SOA(c.g.c)$

$\to \widehat{SMP}=\widehat{SAO}=\widehat{OAP}$

$\to AOMP$ nội tiếp

$\to \widehat{PMS}=\widehat{OAP}=\widehat{OPA}=\widehat{OMA}$

$\to 90^o-\widehat{PMS}=90^o-\widehat{OMA}$

$\to \widehat{BMA}=\widehat{BMP}$

$\to MB$ là phân giác $\widehat{AMP}$

$\to \widehat{AMB}=\dfrac12\widehat{AMP}=\dfrac12\widehat{AOP}=\widehat{ACP}$

$\to \widehat{AMC}=180^o-\widehat{AMB}=180^o-\widehat{ACP}=\widehat{ABP}$

Mà $\widehat{ACM}=\widehat{ACB}=\widehat{APB}$

$\to \Delta AMC\sim\Delta ABP(g.g)$

$\to \widehat{MAC}=\widehat{BAP}$

b.Thiếu $G$