Đáp án+Giải thích các bước giải:

$\Delta OBC$ cân tại $O$, đường cao $OH$

$\Rightarrow OH$ vừa là đường cao vừa là phân giác, trung trực

$\Rightarrow \widehat{O_2}=\widehat{BOA}(1)\\ \widehat{BOA}+\widehat{B_1}=90^\circ\\ \widehat{HBA}+\widehat{B_1}=90^\circ\\ \Rightarrow \widehat{BOA}=\widehat{HBA}(2)\\ (1)(2)\Rightarrow \widehat{O_2}=\widehat{HBA}$

Mà $\widehat{O_1}+\widehat{O_2}=90^\circ$

$\widehat{BAO}+\widehat{HBA}=90^\circ\\ \Rightarrow \widehat{O_1}=\widehat{BAO}\\ \Leftrightarrow \widehat{AOM}=\widehat{BAN}\\ DO \perp OC\\ AC \perp OC\\ \Rightarrow DO//AC\\ \Rightarrow \widehat{O_1}=\widehat{A_1}$

Xét $\Delta CHA$ và $\Delta ADO$

$\widehat{H_1}=\widehat{D_1}=90^\circ\\ \widehat{A_1}=\widehat{O_1}\\ \Rightarrow  \Delta CHA \backsim \Delta ADO\\ \Rightarrow \dfrac{CH}{AD}=\dfrac{HA}{DO}\\ \Leftrightarrow \dfrac{CH}{AD}=\dfrac{2HN}{2DM}\\ \Leftrightarrow \dfrac{CH}{AD}=\dfrac{HN}{DM}$

Xét $\Delta CHN$ và $\Delta ADM$

$\widehat{H_1}=\widehat{D_1}=90^\circ\\ \dfrac{CH}{AD}=\dfrac{HN}{DM}\\ \Rightarrow \Delta CHN \backsim \Delta ADM\\ \Rightarrow \widehat{HNC}=\widehat{DMA}(3)$

$OA$ là trung trực $BA, N \in BC$

$\Rightarrow BN=CN$

$\Rightarrow \Delta CBN$ cân tại $N$

$\Rightarrow \widehat{BNO}=\widehat{CNO} \Leftrightarrow \widehat{BNO}=\widehat{CNH}(4)\\ (3)(4)\Rightarrow \widehat{BNO}=\widehat{DMA}\\ \Rightarrow 180^\circ-\widehat{BNO}=180^\circ-\widehat{DMA}\\ \Leftrightarrow \widehat{BNA}= \widehat{AMO}$

Xét $\Delta BNA$ và $\Delta AMO$

$\widehat{BNA}= \widehat{AMO}\\ \widehat{BAN}=\widehat{AOM}\\ \Rightarrow \Delta BNA \backsim \Delta AMO\\ \Rightarrow \dfrac{BA}{AO}=\dfrac{NA}{MO}\\ \Rightarrow BA.MO=AO.NA.$