$\color{darkgoldenrod}{a)}\ \ \ $Xét $\triangle ABN$ và $\triangle ACP$ có:

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

$\hspace{0.65cm}\ \widehat{ANB}=\widehat{APC}\ (=90^\circ )$

$\Rightarrow\ \triangle ABN\ \backsim\ \triangle ACP\ (g-g)\ (đpcm)$

$\\$

$\Rightarrow\ \dfrac{AB}{AC}=\dfrac{AN}{AP}$

$\Rightarrow\ AB.AP=AC.AN\ (đpcm)$

$\\$$\\$

$\color{darkgoldenrod}{b)}\ \ \ $Xét $\triangle PHB$ và $\triangle NHC$ có:

$\hspace{0.65cm}\ \widehat{PHB}=\widehat{NHC}\ (đối\ đỉnh)$

$\hspace{0.65cm}\ \widehat{BPH}=\widehat{CNH}\ (=90^\circ )$

$\Rightarrow\ \triangle PHB\ \backsim\ \triangle NHC\ (g-g)\ (đpcm)$

$\\$

$\hspace{0.65cm}\ \widehat{BPC}=90^\circ$

$\Rightarrow\ P\in$ đường tròn đường kính $BC$

$\hspace{0.65cm}\ \widehat{BNC}=90^\circ$

$\Rightarrow\ N\in$ đường tròn đường kính $BC$

$\Rightarrow\ BPNC$ là tứ giác nội tiếp

$\hspace{0.65cm}\ $Xét $(BPNC)$ có:

$\hspace{0.65cm}\ \widehat{NPH}$ là góc nội tiếp chắn $\mathop{NC}\limits^{\displaystyle\frown}$

$\hspace{0.65cm}\ \widehat{CBH}$ là góc nội tiếp chắn $\mathop{NC}\limits^{\displaystyle\frown}$

$\Rightarrow\ \widehat{NPH}=\widehat{CBH}\ (đpcm)$

$\\$$\\$

$\color{darkgoldenrod}{c)}\ \ \ BPNC$ là tứ giác nội tiếp $(cmt)$

$\Rightarrow\ \widehat{QPB}=\widehat{QCN}$

$\hspace{0.65cm}\ $Xét $\triangle QPB$ và $\triangle QCN$ có:

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

$\hspace{0.65cm}\ \widehat{QPB}=\widehat{QCN}\ (cmt)$

$\Rightarrow\ \triangle QPB\ \backsim\ \triangle QCN\ (g-g)$

$\Rightarrow\ \dfrac{QB}{QN}=\dfrac{QP}{QC}$

$\Rightarrow\ QB.QC=QN.QP\ (đpcm)$

$\\$$\\$

$\color{darkgoldenrod}{d)}\ \ \ $Dễ c/m: $IKMB,HKME,EFCM$ là các tứ giác nội tiếp

$\Rightarrow\ \begin{cases} \widehat{IKB}=\widehat{IMB}\ \ \ \ \ \color{firebrick}{(1)}\\ \widehat{HKE}=\widehat{HME}\ \ \ \ \ \color{firebrick}{(2)}\\ \widehat{HEK}=\widehat{HMK}\ \ \ \ \ \color{firebrick}{(3)}\\ \widehat{FEC}=\widehat{FMC}\ \ \ \ \ \color{firebrick}{(4)} \end{cases}$

$\hspace{0.65cm}\ $Xét $\triangle HME$ và $\triangle HCM$ có:

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

$\hspace{0.65cm}\ \widehat{HEM}=\widehat{HMC}\ (=90^\circ )$

$\Rightarrow\ \triangle HME\ \backsim\ \triangle HCM\ (g-g)$

$\Rightarrow\ \widehat{HME}=\widehat{HCM}\ \ \ \ \ \color{firebrick}{(5)}$

$\hspace{0.65cm}\ \left.\begin{matrix} MI\bot AB\ (gt)\\CP\bot AB\ (đường\ cao) \end{matrix}\right\}\ \ MI//CP$

$\Rightarrow\ \widehat{IMB}=\widehat{HCM}\ (so\ le\ trong)\ \ \ \ \ \color{firebrick}{(6)}$

$\hspace{0.65cm}\ $Từ $(1),(2),(5),(6)\ \Rightarrow\ \widehat{IKB}=\widehat{HKE}$

$\Rightarrow\ \overline{I,K,E}\ \ \ \ \ \color{firebrick}{(7)}$

$\hspace{0.65cm}\ $Xét $\triangle HMK$ và $\triangle HBM$ có:

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

$\hspace{0.65cm}\ \widehat{HKM}=\widehat{HMB}\ (=90^\circ )$

$\Rightarrow\ \triangle HMK\ \backsim\ \triangle HBM\ (g-g)$

$\Rightarrow\ \widehat{HMK}=\widehat{HBM}\ \ \ \ \ \color{firebrick}{(8)}$

$\hspace{0.65cm}\ \left.\begin{matrix} MF\bot AC\ (gt)\\BN\bot AC\ (đường\ cao) \end{matrix}\right\}\ \ MF//BN$

$\Rightarrow\ \widehat{FMC}=\widehat{HBM}\ (so\ le\ trong)\ \ \ \ \ \color{firebrick}{(9)}$

$\hspace{0.65cm}\ $Từ $(3),(4),(8),(9)\ \Rightarrow\ \widehat{HEK}=\widehat{FEC}$

$\Rightarrow\ \overline{K,E,F}\ \ \ \ \ \color{firebrick}{(10)}$

$\hspace{0.65cm}\ $Từ $(7),(10)\ \Rightarrow\ \overline{I,K,E,F}\ (đpcm)$