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

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

$\hspace{0.65cm}\ $Xét tứ giác $OMNC$ có: $\widehat{DNC}+\widehat{COM}=90^\circ+90^\circ=180^\circ$

$\Rightarrow\ OMNC$ là tứ giác nội tiếp $(đpcm)$

$\\$

$\color{darkgoldenrod}{b)}\ \ $ Xét $\triangle DOM$ và $\triangle DNC$ có:

$\hspace{0.65cm}\ \widehat{CDN}$ chung

$\hspace{0.65cm}\ \widehat{DOM}=\widehat{DNC}\ (=90^\circ)$

$\Rightarrow\ \triangle DOM\ \backsim\ \triangle DNC\ \ (g-g)$

$\Rightarrow\ \dfrac{DO}{DN}=\dfrac{DM}{DC}$

$\Leftrightarrow\ DM.DN=DO.DC=2R.R=2R^2\ \ (đpcm)$

$\\$

$\hspace{0.65cm}\ (CDE)\cap BC=F$

$\Rightarrow\ CEDF$ là tứ giác nội tiếp

$\Rightarrow \begin{cases} \widehat{EFD}=\widehat{ECD}=90^\circ\ \Rightarrow\ EF\bot DF\\ \widehat{EFC}=\widehat{EDC}=\widehat{CBN}\ \Rightarrow\ EF//BN \end{cases}$

$\Rightarrow\ BN\bot DF$

$\hspace{0.65cm}\ $Mà $AN\bot BN\ \ (\widehat{ANB}=90^\circ )$

$\Rightarrow\ DF//AN\ \ (đpcm)$

$\\$

$\color{darkgoldenrod}{c)}$   Xét $\triangle BOC$ vuông tại $O$ có:

$\hspace{0.65cm}\ BC^2=OB^2+OC^2\ \ (Pitago)$

$\Rightarrow\ BC^2=2R^2$

$\Leftrightarrow\ BC=R\sqrt{2}=AC=BD=AD$

$\hspace{0.65cm}\ \triangle DOM\ \backsim\ \triangle DNC\ \ (cmt)$

$\Rightarrow\ \dfrac{OM}{NC}=\dfrac{DM}{DC}$

$\hspace{0.65cm}\ $Xét $\triangle AMN$ và $\triangle DMB$ có:

$\hspace{0.65cm}\ \widehat{MAN}=\widehat{MDB}\ (=\dfrac{1}{2}sđ\mathop{NB}\limits^{\displaystyle\frown})$

$\hspace{0.65cm}\ \widehat{ANM}=\widehat{MBD}\ (=\dfrac{1}{2}sđ\mathop{AD}\limits^{\displaystyle\frown})$

$\Rightarrow\ \triangle AMN\ \backsim\ \triangle DMB\ \ (g-g)$

$\Rightarrow\ \dfrac{AN}{AM}=\dfrac{BD}{DM}$

$\Rightarrow\ \dfrac{OM}{NC}.\ \dfrac{AN}{AM}=\dfrac{DM}{DC}.\ \dfrac{BD}{DM}=\dfrac{BD}{DC}=\dfrac{R\sqrt{2}}{2R}=\dfrac{1}{\sqrt{2}}$

$\Rightarrow\ \dfrac{OM}{AM}=\dfrac{1}{\sqrt{2}}.\ \dfrac{CN}{AN}=\dfrac{CN}{AN\sqrt{2}}$

$\Rightarrow\ \dfrac{OA}{AM}=\dfrac{OM+AM}{AM}=\dfrac{CN+AN\sqrt{2}}{AN\sqrt{2}}\ \ \ \ \ \color{red}{(1)}$

$\hspace{0.65cm}\ $Xét $\triangle BOP$ và $\triangle BNA$ có:

$\hspace{0.65cm}\ \widehat{NBA}$ chung

$\hspace{0.65cm}\ \widehat{BOP}=\widehat{BNA}\ (=90^\circ)$

$\Rightarrow\ \triangle BOP\ \backsim\ \triangle BNA\ \ (g-g)$

$\Rightarrow\ \dfrac{OP}{AN}=\dfrac{BP}{AB}$

$\hspace{0.65cm}\ $Xét $\triangle NCP$ và $\triangle DBP$ có:

$\hspace{0.65cm}\ \widehat{NCP}=\widehat{DBP}\ (=\dfrac{1}{2}sđ\mathop{ND}\limits^{\displaystyle\frown})$

$\hspace{0.65cm}\ \widehat{CNP}=\widehat{BDP}\ (=\dfrac{1}{2}sđ\mathop{BC}\limits^{\displaystyle\frown})$

$\Rightarrow\ \triangle NCP\ \backsim\ \triangle DBP\ \ (g-g)$

$\Rightarrow\ \dfrac{NC}{CP}=\dfrac{BD}{BP}$

$\Rightarrow\ \dfrac{OP}{AN}.\ \dfrac{NC}{CP}=\dfrac{BP}{AB}.\ \dfrac{BD}{BP}=\dfrac{BD}{AB}=\dfrac{R\sqrt{2}}{2R}=\dfrac{1}{\sqrt{2}}$

$\Rightarrow\ \dfrac{OP}{CP}=\dfrac{1}{\sqrt{2}}.\ \dfrac{AN}{CN}=\dfrac{AN}{CN\sqrt{2}}$

$\Rightarrow\ \dfrac{OC}{CP}=\dfrac{OP+CP}{CP}=\dfrac{AN+CN\sqrt{2}}{CN\sqrt{2}}\ \ \ \ \ \color{red}{(2)}$

$\hspace{0.65cm}\ $Từ $\color{red}{(1),(2)}$

$\Rightarrow\ \dfrac{OA}{AM}+\dfrac{OC}{CP}\ge 2\sqrt{\dfrac{OA}{AM}.\ \dfrac{OC}{CP}}$

$\hspace{3.5cm} =2\sqrt{\dfrac{CN+AN\sqrt{2}}{AN\sqrt{2}}.\ \dfrac{AN+CN\sqrt{2}}{CN\sqrt{2}}}$

$\hspace{3.5cm} =2\sqrt{\dfrac{(CN+AN\sqrt{2})(AN+CN\sqrt{2})}{2AN.CN}}$

$\hspace{3.5cm} =2\sqrt{\dfrac{3AN.CN+\sqrt{2}(AN^2+CN^2)}{2AN.CN}}$

$\hspace{3.5cm} \ge 2\sqrt{\dfrac{3AN.CN+\sqrt{2}.2AN.CN}{2AN.CN}}$

$\hspace{3.5cm} =2\sqrt{\dfrac{3+2\sqrt{2}}{2}}$

$\hspace{3.5cm} =\sqrt{6+4\sqrt{2}}$

$\hspace{0.65cm}\ $Đẳng thức xảy ra khi:

$\hspace{0.65cm}\ \begin{cases} AN=CN\\ \dfrac{CN+AN\sqrt{2}}{AN\sqrt{2}}=\dfrac{AN+CN\sqrt{2}}{CN\sqrt{2}}\ \Leftrightarrow\ AN=CN \end{cases}$

$\Rightarrow\ N$ là điểm chính giữa của cung $AC$