it is not hard to observe desired point will be $A$-$Dumpty$ with angle it is okay. So we want to prove $D_{A},O,D$ Collinear Let's take $OD \cap {AD_{A}}$ at $D_{A}'$ with angles it will be easy contradiction

We complex bash with $\omega$ as the unit circle, so that $$|a|=|b|=|c|=1$$$$o=0$$$$m=\frac{b+c}2$$$$d=\frac{a(ab+ac-2bc)}{a^2-bc}$$$$k = b + m - ab\overline{m} = \frac{c^2+3bc-ab-ac}{2c}$$$$\ell = c + m - ac\overline{m} = \frac{b^2+3bc-ab-ac}{2b}$$Now we want to find the coordinate of $X$. Since $BX\parallel KM$, we have $$\frac{b-x}{k-m} \in \mathbb{R}$$$$\frac{2c(b-x)}{2bc-ab-ac} = \frac{2a(1-b\overline{x})}{2a-b-c}$$$$c(b-x)(2a-b-c) = a(1-b\overline{x})(2bc-ab-ac)$$$$\overline{x} = \frac{(b+c)(a^2-bc) - cx(2a-b-c)}{ab(ab+ac-2bc)}$$Similarly, since $CX\parallel LM$, we have $$\overline{x} = \frac{(b+c)(a^2-bc) - bx(2a-b-c)}{ac(ab+ac-2bc)}$$We set these equal to get $$\frac{(b+c)(a^2-bc) - cx(2a-b-c)}{ab(ab+ac-2bc)} = \frac{(b+c)(a^2-bc) - bx(2a-b-c)}{ac(ab+ac-2bc)}$$$$c(b+c)(a^2-bc) - c^2x(2a-b-c) = b(b+c)(a^2-bc) - b^2x(2a-b-c)$$$$x(b+c)(b-c)(2a-b-c) = (a^2-bc)(b+c)(b-c)$$$$x = \frac{a^2-bc}{2a-b-c}$$Then $$\frac{x-o}{d-o} = \frac{(a^2-bc)^2}{a(ab+ac-2bc)(2a-b-c)} = \frac{|a^2-bc|^2}{|2a-b-c|^2}$$which is real. $\blacksquare$

Redefine $X$ as the intersection of the circle passing through $A$ and $B$ that is tangent to $AC$ and the circle passing through $A$ and $C$ that is tangent to $BC$. Claim: $BX||KM$ and $CX||LM$ By $\sqrt{bc}$ inversion circle $(ACX)$ is mapped to the line through $B$ parallel to $AC$ and $(ABX)$ is mapped to the line through $C$ parallel to $AB$ thus $X$ is mapped to the reflection of $A$ across $M$. This implies that $AX$ is a symmedian. Left is in easy angle chase $$\angle AKM=\angle BAM=\angle XAC=\angle XBA$$$$\angle ALM=\angle CAM=\angle XAB=\angle XCA$$ Claim: $D$ and $X$ are inverses with respect to $(ABC)$ It is sufficient to show that $(ADB)$ and $(AXB)$ are inverses and $(ADC)$ and $(AXC)$ are inverses. Notice that $(ADB)$ is mapped to the circle passing through $A$ and $B$ and tangent to $(AOC)$ and $(ADC)$ is mapped to a circle passing through $A$ and $C$ and tangent to $(AOB)$. Thus we need to show that $(AOC)$ and $(ADB)$ are tangent and $(AOB)$ and $(ADC)$ are tangent. $$\angle OCA+\angle BDA=\angle BAO$$$$\angle OBA+\angle CDA=\angle CAO$$

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Really easy but OP should probably not give spoilers on the problem in the heading but here its not really an issue. The entirety of the solution is the following result. Claim : Point $X$ is the $A-$Dumpty Point of $\triangle ABC$. Proof : First, note that \[\measuredangle CXB = \measuredangle LMK = 2\measuredangle LAK = 2\measuredangle CAB = \measuredangle COB \]from which it follows that $X$ lies on $(BOC)$. Further, \[\measuredangle ABX = \measuredangle AKM = \measuredangle MAK = \measuredangle MAB \]But the $A-$Dumpty point of $\triangle ABC$ is well known to satify this angle condition as well. Since we know the Dumpty Point lies on $(BOC)$ as well, it follows that $X$ is indeed the $A-$Dumpty Point of $\triangle ABC$ as claimed. Now, note that this means $X$ lies on the circle with diameter $AO$. Now, by Radical Center on circles $(AO)$ , $(ABC)$ and $(BOC)$, it follows that the tangent to $(ABC)$ at $A$ , $XO$ and $BC$ concur. Since $D$ is the intersection of the tangent to $(ABC)$ at $A$ and $BC$ , it follows that $D-X-O$ as desired.

Invert wrt $\omega$. Then $D',M'\in (BOC)$ because $B,C,D,M$ are collinear and $A,D',M'$ are collinear because $AOMD$ is cyclic. Hence\begin{align*}\angle D'BA & =\angle OBA-\angle OBD' \\ & =\angle BAO-\angle OBD' \\ & =\angle BAO-\angle OM'D' \\ & =\angle BAO-\angle OM'A \\ & =\angle BAO-\angle MAO \\ & =\angle BAM \\ & =\angle KAM \\ & =\angle MKA, \end{align*}so $BD'\parallel KM$. Analogously we get $CD'\parallel LM$. Hence $X=D'$, so $X,D,O$ are collinear.

Prove that XOCB is cyclic then prove angleDOB= angle XCB