Q:

Is $\phi =\angle A"OB" = \measuredangle(AB,A"B")=\measuredangle(A'B',A"B")$? [Doubt]

Can someone clarify this doubt ?
We denote the spiral similarity by $S$, the rotation centered at $O$  with
angle $\phi$ by $\rho _O ,\phi$ , and the homothety centered at $O$ with ratio $k$ by $\chi _{ O, k}$ , then
$S _{O, k, \phi}$ = $\rho_O, \phi \circ \chi _{ O, k}$ .
Consider the following image :

We are given a triangle $ABC$ which is being dilated by a spiral symmetry $S$ centred at $O$ with ratio $k$ and and angle $\phi$ .
I noted that since angles are preserved in dilation and homothety, we get that $\Delta ABC \sim \Delta A"B"C" $.
Also $\Delta OAB \sim \Delta OA"B"$. And we have $\angle A"OB"=\angle AOB$ .
And we also have $\measuredangle(AB,A"B")=\measuredangle(A'B',A"B")$ (since $A'B'||AB$) .
But I couldn't understand how $\measuredangle(AB,A"B")=\measuredangle(A'B',A"B")=\phi$ . Isn't $\phi =\angle A"OB"$ ?
Here is the whole explanation of the book: 

A:

It should be $\phi = \angle A'OA''$.
Rotation takes $A'$ to $A''$ and $B'$ to $B''$ so it takes $A'B'$ to $A''B''$ so we have $$\angle (A''B'',A'B') = \phi$$
Since homothety takes $A$ to $A'$ and $B$ to $B'$ we have $AB||A'B'$, so it is obivously that $$\angle (A''B'',A'B') =\angle (A''B'',AB)$$

