Giải thích các bước giải:

a.Vì $\widehat{BFC}=\widehat{BEC}=90^o$

$\to BCEF$ nội tiếp

$\to \widehat{FEA}=\widehat{ABC}$

Ta có: $\Delta BEC$ vuông tại $E, M$ là trung điểm $BC$

$\to ME=MB=MC=\dfrac12BC$

$\to \Delta MEC$ cân tại $M$

$\to \widehat{MEC}=\widehat{MCE}=\widehat{ACB}$

Ta có:

$\widehat{NBM}=180^o-\widehat{SBC}=180^o-\widehat{BAC}=\widehat{ABC}+\widehat{ACB}=\widehat{AEF}+\widehat{MEC}=180^o-\widehat{NEM}$

$\to \widehat{NBM}+\widehat{NEM}=180^o$

$\to MBNE$ nội tiếp
$\to \widehat{NMB}=\widehat{NEB}=\widehat{BEF}=ư\widehat{BCF}$

$\to MN//CF$

Do $CF\perp AB$

$\to MN\perp BF$

b.Xét $\Delta SBP,\Delta SAB$ có:

Chung $\hat S$

$\widehat{SBP}=\widehat{SAB}$

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

$\to \dfrac{BP}{AB}=\dfrac{SB}{SA}$

Tương tự: $\dfrac{CP}{AC}=\dfrac{SC}{SA}$

Mà $SA, SB$ là tiếp tuyến của $(O)$

$\to SB=SC$

$\to \dfrac{BP}{AB}=\dfrac{CP}{AC}$

$\to AB.CP=AC.BP$

c.Từ b $\to \dfrac{SB}{SP}=\dfrac{SA}{SB}$

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

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

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

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

$\to AOMP$ nội tiếp

$\to \widehat{AMO}=\widehat{APO}=\widehat{OAP}=\widehat{PMS}$

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

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

$\to MB$ là phân giác $\widehat{AMP}$

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

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

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

$\to \Delta ABP\sim\Delta AMC(g.g0$

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