Theorem reference: inscribed angle bisector Given a circle $\omega$, let $O,A,B,K \in \omega$ (so that $OAKB$ is cyclic quadrilateral) and $\angle KOA = \angle KOB = \phi$.
I found that $OA + OB = 2 OK \cos \phi$. Is there a well-known theorem (something like power of point) which implies that?
I got the result using similarity and law of cosines (which is ugly).
Any other result on $OA + OB = (...)$ will be fine.
UPD:
I guess I need to clarify my motives at this point.
What I originally had in mind is the following. Given a bisector $OK$ of an angle $\angle AOB$ (with fixed points $A,B,K$) what is the condition for points $O,A,K,B$ to be concyclic? After some manipulation with complex numbers, I got the result I originally mentioned. So originally I hopped that power of point $O$ of some sort may appear (note that if $O$ was out of circle, concyclic condition is exactly the power of point theorem).
 A: Thanks to @ACB I finally found what I was looking for (just didn't see it quite well). Using Ptolemy's theorem and knowing that $AK = BK$ we get
$$
OA \cdot BK + OB \cdot AK = OK \cdot AB \Rightarrow OA + OB = \frac{AB}{AK} \cdot OK
$$
Now we can find that $\frac{AB}{AK} = 2 \cos \phi$ using similarity or other way around (e.g. complex numbers as I did it originally).
Thanks to everyone else as well!
UPD:
Check the @Blue comment to get a more generic approach:
$$
OA \sin \angle KOB + OB \sin \angle KOA = OK \sin \angle AOB
$$
A: An approximate construction as an aid to proof when the bisecting diagonal is not the diameter of the circle.
Angles subtended at center, circumference are $60^{0},30^{0}$
Central diagonal $OK$ length = $19.57; 19.57 \cos 30^{0}=16.95$
Average of equal angle enclosing arms $14.89,19.01 = 16.95$
For the proof we consider the polar equation of a circle passing through the origin. Angle subtended at rim is half that subtended at circle center.
$$OA= \rho_1= d \cos \alpha,~OK=\rho_2= d \cos (  \alpha+ \theta),~ OB=\rho_3= d \cos ( \alpha+ 2\theta); $$
Sum of two cosines trig formula
$$ \frac {\rho_1+\rho_3}{2} = d \cos(\alpha +\theta) \cos \theta  $$
$$ \rho_2 \cos \theta  = d \cos(\alpha +\theta) \cos \theta. $$
This is chord length expression rather than a theorem for an average/ middle polar angle argument.
$$ \frac {\rho_1+\rho_3}{2} = d \cos(\alpha +\theta) \cos \theta  $$
$$ \rho_2 \cos \theta  = d \cos(\alpha +\theta) \cos \theta. $$
Towards a generalization or theorem, not yet checked :
A property of the circle containing uniformly spaced pencil of chords (i.e., with uniform angular separation emanating from a point on the circumference) may be stated. If
$$ \bar{\rho}= \frac {\sum_{i=1
}^n \rho_i} {n}; ~ \bar{\theta}= \frac {\sum_{i=1
}^n \theta} {n}; $$
then
$$  \bar{\rho}=d \cos  \bar{\theta}\cos \theta. $$

A: Let $\angle OKA=x$, $\angle OKB=y$, $\angle OBK=B$, $\angle OAK=A$ and the radius of (circum)circle be $R$.
From extended law of sines, $OA=2R\sin x$, $OB=2R\sin y$ and $OK=2R\sin A$.
On the other hand, in $\triangle OKA$, since $A+\phi=180^{\circ}-x$ $$\sin x=\sin A\cos\phi+\cos A\sin\phi\tag1$$ and in $\triangle OKB$, $\sin y=\sin(B+\phi)=\sin B\cos\phi+\cos B\sin\phi$, and by using $\cos B=-\cos A$ and $\sin B=\sin A$, $$\sin y=\sin A\cos\phi-\cos A\sin\phi\tag2$$
From $(1)$ and $(2)$, $$\sin x +\sin y=2\cos\phi\sin A.\tag3$$
By multiplying $(3)$ by $2R$, we get $OA+OB=2OK\cos\phi.$
