My solution: Let $H_{\infty}, I_{\infty}$ be the infinity point on $AA', AD$, respectively . Let $T=AA' \cap ON$ and $S, R$ be the midpoint of $AO, AD$, respectively . Easy to see $M, S, R$ are collinear . Since $\angle ATO=\angle XAD=\angle DAO=\angle AOT$ , so we get $AT=AO=DO \Longrightarrow ATOD$ is a parallelogram $\Longrightarrow T, S, D$ are collinear . Notice that $\{AA', AO \}$ are isogonal conjugate of $\angle BAC$ , so from $A(M,O;D,A')=D(M,S;R,H_{\infty})=(N,T;I_{\infty},O)=A(N,A';D,O) \Longrightarrow \angle BAM=\angle CAN$ . Q.E.D

If $AM$ cuts $OD$ at $P,$ then $AXPD$ is obviously parallellogram.Thus, from $\triangle PDX \cong \triangle AXD \sim \triangle ODN,$ we get $\tfrac{AX}{XP}=\tfrac{AX}{AD}=\tfrac{OD}{ON}=\tfrac{AO}{ON}.$ But clearly $\angle AXP=\angle AON$ $\Longrightarrow$ $\triangle AXP \sim \triangle AON$ by SAS $\Longrightarrow$ $\angle XAM=\angle OAN$ $\Longrightarrow$ $AM,AN$ are isogonals WRT $AO,AX,$ which in turn are isogonals WRT $\angle BAC$ $\Longrightarrow$ $\angle BAM=\angle CAN.$

Enough to show that $\angle XAM = \angle NAO$, for $AA'$ and $AO$ are symmetrical to $AD$. $M'$ is the midpoint of $DN$, and $AD$ and $OM'$ intersect at $X$. 1) $DT$ and $ON$ are parallel, $OT$ bisects $DN$ -> $OD$ and $NT$ are also parallel. 2) $\angle ATN = \angle DON = \angle A'AD = \angle DAO$ -> $A, O, N, T$ are concyclic $XAD$ and $DON$ are similar -> $\angle XAM = \angle DOM' = \angle OTN = \angle OAN$.

My solution (very easy) Let the midpoint of $DN$ be $L$ and Let $AD \cap OL$ be $K$ then,$ODKN$ is a parallellogram. Now, $AONK$ is an isosceles trapezoid And since $\bigtriangleup AXD \sim \bigtriangleup ODN$ we are done by chasing angles.

ComplexPhi wrote: Let $ABC$ be a triangle, let $O$ be its circumcenter, let $A'$ be the orthogonal projection of $A$ on the line $BC$, and let $X$ be a point on the open ray $AA'$ emanating from $A$. The internal bisectrix of the angle $BAC$ meets the circumcircle of $ABC$ again at $D$. Let $M$ be the midpoint of the segment $DX$. The line through $O$ and parallel to the line $AD$ meets the line $DX$ at $N$. Prove that the angles $BAM$ and $CAN$ are equal. Construct parallelogram $AXPD$. Then $\overline{XP} \parallel \overline{AD} \parallel \overline{ON}$ so $\overline{XP} \cap \overline{ON}$ and $\overline{XN} \cap \overline{PO}$ are isogonal in angle $OAX$. Thus, by converse of isogonality lemma; we see $\angle PAX=\angle OAN$ hence $\angle BAM=\angle CAN$ as desired. $\blacksquare$

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This turned out to be a lot easier than expected. ComplexPhi wrote: Let $ABC$ be a triangle, let $O$ be its circumcenter, let $A'$ be the orthogonal projection of $A$ on the line $BC$, and let $X$ be a point on the open ray $AA'$ emanating from $A$. The internal bisectrix of the angle $BAC$ meets the circumcircle of $ABC$ again at $D$. Let $M$ be the midpoint of the segment $DX$. The line through $O$ and parallel to the line $AD$ meets the line $DX$ at $N$. Prove that the angles $BAM$ and $CAN$ are equal. Clearly, we need to show that $\measuredangle XAM=\measuredangle NAO.$ Let $P$ be the reflection of $A$ over $M$ so that $AXPD$ is a parallelogram. Now we present the key claim. Note that the result is directly implied by the claim: Claim We have $\triangle AXP \overset{-}{\sim} \triangle AON.$ Proof: Firstly, $\measuredangle AXP=\measuredangle ODA=\measuredangle DAO=-\measuredangle AON.$ Next, since $AX \parallel OD$ and $AD \parallel ON,$ we get $\triangle AXD \overset{+}{\sim} \triangle ODN$ so $$ON: AD=OD:AX=OA:AX \implies ON:OA=AD:AX=XP:AX$$ Hence $\triangle AXP \overset{-}{\sim} \triangle AON.$ $\square$

ComplexPhi wrote: Let $ABC$ be a triangle, let $O$ be its circumcenter, let $A'$ be the orthogonal projection of $A$ on the line $BC$, and let $X$ be a point on the open ray $AA'$ emanating from $A$. The internal bisectrix of the angle $BAC$ meets the circumcircle of $ABC$ again at $D$. Let $M$ be the midpoint of the segment $DX$. The line through $O$ and parallel to the line $AD$ meets the line $DX$ at $N$. Prove that the angles $BAM$ and $CAN$ are equal. Solution: WLOG, Assume, $N$ lies outside $\odot(ABC)$. Let $E$ be midpoint of $AD$, $D'$ be reflection of $D$ over $O$ and $G$ be midpoint of $AD'$. Hence, $G \in EM$, $GM||DD'$ and $O \in GN$. Apply Pappus' Theorem on $DAMGND'$ $\implies$ $AM||ND'$ $\implies$ $\angle DAN$ $=$ $\angle ANG$ $=$ $\frac12 \angle AND'$ $=$ $\frac12 \angle MAN$ $\qquad \blacksquare$

Let $P$ be the intersection of the external bisector of angle $BAC$ and circumcircle and $Q$ be the midpoint of $AP$. Let $S$ be the intersection of $AQ$ and $XD$. It suffices to show that $-1=(M, N;D, S)$.Since $AD \perp AS$ it directly implies $\angle MAD = \angle DAN$. First, notice $QM \parallel AX \parallel PD$ Now we will move harmonic quadruples by taking $Q$ center. $-1 = (P,D;O,P_{\infty})=(S,D;N,M)=(M,N;D,S)$ and we are done.