$\color{darkgoldenrod}{a)}\ \ \ \widehat{ACB}$ là góc nội tiếp chắn nửa đường tròn $(O)$

$\Rightarrow\ \widehat{ACB}=90^\circ$

$\Rightarrow\ AC\bot BC$

$\hspace{0.65cm}\ $Mà $OH\bot AC\ (gt)$

$\Rightarrow\ OH//BC\ (từ\ \bot\ đến\ //)\ (đpcm)$

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$\color{darkgoldenrod}{b)}\ \ \ OA=OC=R$

$\Rightarrow\ \triangle OAC$ cân tại $O$

$\hspace{0.65cm}\ $Xét $\triangle OAC$ cân tại $O$ có $OH$ là đường cao $(OH\bot AC)$

$\Rightarrow\ OH$ là phân giác $\widehat{AOC}$

$\Rightarrow\ \widehat{AOM}=\widehat{COM}$

$\hspace{0.65cm}\ $Xét $\triangle AOM$ và $\triangle COM$ có:

$\hspace{0.65cm}\ OM$ chung

$\hspace{0.65cm}\ \widehat{AOM}=\widehat{COM}\ (cmt)$

$\hspace{0.65cm}\ OA=OC=R$

$\Rightarrow\ \triangle AOM=\triangle COM\ (c-g-c)$

$\Rightarrow\ \widehat{OAM}=\widehat{OCM}=90^\circ$

$\Rightarrow\ AM\bot AB\ \ (đpcm)$

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$\color{darkgoldenrod}{c)}\ \ \ OH//BC\ (cmt)$

$\Rightarrow\ \widehat{AOM}=\widehat{KBC}\ (so\ le\ trong)$

$\hspace{0.65cm}\ $Xét $\triangle MAO$ và $\triangle CKB$ có:

$\hspace{0.65cm}\ \widehat{AOM}=\widehat{KBC}\ (cmt)$

$\hspace{0.65cm}\ \widehat{MAO}=\widehat{CKB}\ (=90^\circ)$

$\Rightarrow\ \triangle MAO\ \backsim\ \triangle CKB\ (g-g)\ \ (đpcm)$