Puzzlement about a step in WW Rouse Ball's explanation of Hero's proof I have a 4th edition copy of "A Short Account of the History of Mathematics" by W. W. Rouse Ball.  This work, written in the late 1800's (the 4th edition dates to the early 1900's) is a highly respected history, and contains accessible explanations of many historically important results.
In his explanation of how Hero derived the formula $\sqrt{s(s-a)(s-b)(s-c)}$ for the area of a triangle with sides $a, b, c$ and semi-perimeter $s$, on page 90 in chapter 4, Ball sets up a triangle $ABC$ with inscribed circle centered at $O$, touching side $AB$ at point $F$.  He then defines point K as the meeting of two lines:  The line passing through $C$ perpendicular to side $BC$, and the line passing through $O$ perpendicular to $BO$.  Then the following statement is made:

He then shews (sic) that the angle $OAF = $ angle  $CBK$; hence the triangles $OAF$ and $CBK$ are similar.

(Note that "shews" rhymes with "sews," not with "news.")  The formula then follows from a sequence of ratio manipulations.
I cannot prove or even convince myself that those two angles are equal.  I don't restrict the proof to classical geometry; I have tried the usual vector tricks (which for example easily demonstrate that if side $BC = \vec{u}$ and $BA = \vec{v}$ the angle bisectors meet at a point which differs from $B$ by $\frac{\vec{u}+\vec{v}}{3}$).  I have also tried using analytic geometry; the coordinates of $K$ get pretty messy and I was unable to "shew" that the two discussed angles are equal.
Could it be that this respected author simply erred?  I doubt it, because the subsequent ratio manipulations would then hint that Hero's formula does not work.
So I am looking for any simple demonstration that those two angles are equal.
 A: As @YNK observes, the key is that, because $O$ and $C$ subtend equal angles with $\overline{BK}$, we have $\square OBKC$ is cyclic. Here's an alternative angle chase to the result:

$$\angle CBK \underbrace{=}_{\text{Insc}\angle\text{Thm}} \angle COK \underbrace{\;=\;}_{\triangle OBC} 180^\circ - \left(\tfrac12B+\tfrac12C+90^\circ\right)=\tfrac12(180^\circ-B-C)=\tfrac12 A$$
A: 
Let $\measuredangle CAB$, $\measuredangle ABC$, and $\measuredangle BCA$ be $2\alpha$, $2\beta$, and $2\omega$ respectively. For brevity, we also denote $\measuredangle CBK$ as $\phi$.
Since $O$ is the incenter, $OA$, $OB$, and $OC$ are the angle bisectors of $\measuredangle CAB$, $\measuredangle ABC$, and $\measuredangle BCA$ respectively
Since $\measuredangle KOB = \measuredangle KCB = 90^o$, $OBKC$ is a cyclic quadrilateral. Therefore,
$$\measuredangle BKO = \measuredangle BCO = \omega. $$
$\Delta KOB$ is a right angle triangle. Therefore,
$$\measuredangle BKO + \measuredangle OBK =\omega+\beta+\phi = 90^o\quad\rightarrow\quad \phi=90^o -\beta - \omega. \tag{1}$$
Now consider the triangle $ABC$. There we have,
$$2\left(\alpha+\beta+\omega\right)=180^o \quad\rightarrow\quad \alpha=90^o -\beta - \omega. \tag{2}$$
From (1) and (2), it is obvious that $\phi =\alpha$. Therefore,
$$\measuredangle CBK = \measuredangle OAF. $$
