Giải giúp câu b bài 4 với

Giải thích các bước giải:

1.Ta có: $\widehat{BNC}=\widehat{BMC}=90^o$

$\to B, N, M, C\in$ đường tròn đường kính $BC$

2.Kẻ tiếp tuyến $At$ của $(O)$

$\to \widehat{tAB}=\widehat{ACB}=\widehat{BCN}=\widehat{AMN}$

$\to At//MN$

Mà $OA\perp At\to OA\perp MN$

Lại có: $MN\perp MK$

$\to AO//NK$

3.Ta có:

$\widehat{NKM}=90^o-\widehat{NMK}=90^o-\widehat{NMC}=90^o-\widehat{NBC}=\widehat{NCB}=\widehat{NCD}$

$\to \widehat{DHN}=180^o-\widehat{NCD}=180^o-\widehat{HKN}=\widehat{CKN}$

Xét $\Delta NBC,\Delta NMK$ có:

$\widehat{BNC}=\widehat{MNK}$

$\widehat{NMK}=\widehat{NMC}=\widehat{NBC}$

$\to \Delta NMK\sim\Delta NBC(g.g)$

$\to \dfrac{NM}{NB}=\dfrac{MK}{BC}$

$\to MN.BC=KM.NB$

Xét $\Delta NMB,\Delta NKC$ có:

$\widehat{NBM}=\widehat{NCM}=\widehat{NCK}$

$\widehat{NMB}=180^o-\widehat{NCB}=180^o-\widehat{NCD}=\widehat{NHD}=\widehat{NKC}$

$\to \Delta NMB\sim\Delta NKC(g.g)$

$\to \dfrac{MB}{KC}=\dfrac{NB}{NC}$

$\to MB.NC=KC.NB$

$\to BM.CN+MN.BC=KC.NB+MK.NB=NB.MC$