Bump on cartesian plane.

Let $AK$ and $DO$ intersect $(O)$ again at $T$ and $H$ respectively. Let $T'$ be the antipode of $T$. We have $AT'\parallel MB$ and $DB\parallel MN$. So, $$\angle{CDT'}=\angle{CDB}+\angle{BMT'}=\angle{CDB}+\angle{AT'M}=\angle{CDB}+\angle{MNC}=180^{\circ}-\angle{DCN},$$implying $DT'\parallel CN\implies TH\parallel CN$. Pascal's theorem with six points $N,A,C,T,H,D$ finishes the problem.

Since $AC \perp BD$, it suffices to prove that $\angle KAM = \angle CBD$. On one hand, $GK \parallel NC$ implies that $AGKD$ are concyclic. Let $P =GD \cap NC$, then $$\angle KAD =\angle KGD = \angle NPD.$$On the other hand, $\angle DAM = \angle DNM =\angle DNO = \angle ODN$. Thus \[ \angle KAM = \angle KAD + \angle DAM = \angle NPD + \angle ODN = \angle CND =\angle CBD. \]

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Let $\angle A D C=\angle G N C=\theta$. $NC \parallel GK$ $\Rightarrow$ $A,D,K,G$ are concyclic. $N$ is the mid point of arc $ANC$ $\Rightarrow$ $\angle N A C=$ $\angle N C A=$ $\frac{\theta}{2}$. Let $AC$ and $MN$ intersect at $L$. So $\angle A L N=90^{\circ}$ $\Rightarrow$ $BD \parallel MN$ $\Rightarrow$ $\angle B A N=$ $\angle B M N=$ $\angle D B M=$ $x^{\circ}$. Now $\angle A B M=$ $90-\frac{\theta}{2}$. $\angle A B D=$ $\angle A B M-$ $\angle D B M=$ $90-\frac{\theta}{2}-x=$$\angle A C D$. Now $\angle D O A=$ $2 \angle D C A=$ $180^{\circ}-\theta-2 x$ $\Rightarrow$ $\angle G D A=$ $\angle O D A=$ $\frac{\theta}{2}+x =$ $\angle A K G$. So, $\angle G A K=$ $\frac{\theta}{2}-x$ $\Rightarrow$ $\angle A B M$ $+\angle B A K$ $=\angle A B M$ $+\angle G A K$ $+\angle B A N = 90^{\circ}$ $\Rightarrow$ $AK \perp BM$. Edit: Sorry for the bad formatting I'm bad in latex

Essentially, the same idea as Flash_Sloth. Look at the following observations and constructions: 1) Let $DG \cap HC = P$, $AC \cap BD = X$ and $AK \cap MB = Y$ 2) By definition, we have that $M,O,N$ are colinear. 3) To prove $\angle AYB = 90^\circ$, it suffices to show that $ABXY$ is cyclic. Now this can be done by showing that $\angle KAM = \angle DBC$, because we already know that $\angle MAC = \angle MBC$. $\textcolor{red}{Claim} :$ $A,D,K$ and $G$ are concylic $\textcolor{blue}{Proof} :$ We already know that $ADCN$ is cyclic. Now since $GK \parallel NC$, we get that : $$\angle DAN = 180^\circ - \angle DCN = 180^\circ - \angle DKG$$Which is basically the desired claim. Now observe that : $$\angle DAK = \angle DGK = \angle DPN$$also note that since $AMDN$ is cyclic and $\Delta DON$ is isosceles, we get : $$\angle MAD = \angle MND = \angle ODN$$Now finally keeping in mind that $DNBC$ is cyclic we get : $$\angle KAM = \angle MAD + \angle KAD = \angle ODN + \angle DPN = \angle DNC = \angle DBC$$And we are done! $\blacksquare$

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