Angle chasing to show three points are collinear. Let $ABC$ be an acute triangle with circumcenter $O$ and let $K$ be such that $KA$ is tangent to
the circumcircle of $\triangle ABC$ and $\angle KCB = 90 ^{\circ}$. Point $ D$ lies on $ BC$ such that $KD || AB.$ Show
that $DO$ passes through $A.$

This is a problem from EGMO, eg.1.32.
My approach:
I created a dummy point D' such that D'O passes through A.
Now I just have to show that BA and D'A are || which will solve the problem. I think I have all the necessary theorems and I think this problem can be solved with angle chasing, but I can't really figure out how to proceed.
Can you please guide me?
 A: We shall try to work backwards in order to reach the desired conclusion.

Let $\triangle ABC$ be an acute triangle with circumcenter $O$ and let $K$ be such that $KA$ is tangent to the circumcircle of $\triangle ABC$ and $\angle KCB=90^{\circ}.$

Define $L$ to be any point on line $KA$ such that $A$ lies between $K$ and $L$. Also, extend segment $AO$ to point $X$ on $BC$.
Using Alternate Segment Theorem,
$$\angle BAL=ACB=\alpha\implies \angle AOB=2\alpha\implies \angle OAB=\angle BAX=90-\alpha. $$
Since, $\angle XAL=\angle XCK=90^{\circ}, AKCX$ is cyclic and hence, $$\angle AKX=\angle ACX=\alpha\implies AB\parallel KX. $$
But there exists a unique point $D$ on $BC$ such that $AB\parallel KD$. Therefore, $X\equiv D$ and $A, O, D$ are collinear.
A: It is sufficient to show that $\angle CAO=\angle CAD$.
Let the tangent be labelled $KAK'$.
Then due to the Alternate Segment Theorem, $$\angle K'AB=\angle ACB$$
And since $AB\parallel KC$, both of these equal $\angle AKD$.
The tangent is perpendicular to the radius so $\angle BAO=90^o-\angle K'AB$.
But from the data given, $$\angle KCA=90^o-\angle ACB\implies \angle KCA=\angle BAO$$
Considering the total angle in triangles $ABC$ and $AKC$, the remaining angle in each is the same, so $$\angle OAC=\angle DKC$$
However, points $AKC$ and $D$ are concyclic since $\angle AKD=\angle ACD$ (due to Angles in the Same Segment are Equal).
Therefore, for the same reason, $\angle DAC=\angle DKC$.
Hence, finally, $$\angle CAO=\angle CAD$$  QED
