Giải thích các bước giải:

a.Ta có:

$\widehat{MEA}=\widehat{MFA}=90^o$

$\to A, M, E, F\in$ đường tròn đường kính $AM$

b.Ta có:

$\widehat{KBC}=\widehat{KAC}=\widehat{MAF}=\widehat{MEF}$

$\widehat{BKC}=180^o-\widehat{BAC}=180^o-\widehat{EAF}=\widehat{EMF}$

$\to \Delta KBC\sim\Delta MEF(g.g)$

$\to\dfrac{KB}{ME}=\dfrac{KC}{MF}$

$\to \dfrac{KB}{KC}=\dfrac{ME}{MF}$

Vì $M$ là trung điểm $BC\to MB=MC$

$\to \dfrac{KB}{KC}=\dfrac{ME}{MB}.\dfrac{MC}{MF}$

$\to KB.\dfrac{MF}{MC}=KC.\dfrac{ME}{MB}$

$\to KB.\sin\widehat{MCF}=KC.\sin\widehat{MBE}$

$\to KB.\sin\widehat{ACB}=KC.\sin\widehat{ABC}$

c.Ta có:

$\Delta KBC\sim\Delta MEF$

$M, Q$ là trung điểm $BC, EF$

$\to \Delta MEQ\sim\Delta KBM, \Delta MFQ\sim\Delta KCM$

$\to \widehat{EMQ}=\widehat{BKM}$

$\to \widehat{EMQ}=\widehat{BKA}$

$\to \widehat{EMQ}-\widehat{EMA}=\widehat{BKA}-\widehat{EMA}$

$\to \widehat{AMQ}=\widehat{BCA}-\widehat{EFA}$

Ta có:

$\widehat{OAM}=\widehat{OAK}=\widehat{OKA}=\widehat{AKB}-\widehat{OKB}$

Mà $\widehat{OKB}=90^o-\dfrac12\widehat{BOK}=90^o-\widehat{BAK}=90^o-\widehat{EAM}=\widehat{AME}=\widehat{AFE}$

$\to \widehat{OAM}=\widehat{ACB}-\widehat{EFA}$

$\to \widehat{AMQ}=\widehat{OAM}$

$\to AO//MQ$