Problem with Directed Angles which sum to ${\pi \over 2}$. I was solving a section in my book(EGMO Lemma 1.30) where the author discusses the uses of directed angles, when I came across-

Points $A, B, C$ lie on a circle with center $O$. Show that $\measuredangle$ $OAC$ = $90^\circ$ − $\measuredangle$ $CBA$.

Let me denote directed angle with $\measuredangle$.(everywhere)

Here is an attempt; the author talks about the directed angles in blue, and it is to be shown that they sum to half $\pi$ radians. Lines in red are my own construction.
By directed angles, we know that $\measuredangle$ $CBA$ = $\measuredangle$ $CXA$ = ${1\over 2}$ $\measuredangle$ $COA$ (the inscribed angle theorem).
And also that $\measuredangle$ $OAC$ = $\measuredangle$ $ACO$ (triangle $OAC$ is isosceles).
Now by a theorem of directed angles, $\measuredangle$ $OAC$ $+$ $\measuredangle$ $ACO$ $+$ $\measuredangle$ $COA=0$
But after this, as we're working modulo $\pi$ radians, it's unintelligible to multiply or divide by $2$, which I have to do, so my attempt failed.
Answers are thankfully welcome.
By directed angles, we know that $\measuredangle$ $CBA$ = $\measuredangle$ $CXA$ = ${1\over 2}$ $\measuredangle$ $COA$ (the inscribed angle theorem).
And also that $\measuredangle$ $OAC$ = $\measuredangle$ $ACO$ (triangle $OAC$ is isosceles).


Now by a theorem of directed angles, $\measuredangle$ $OAC$ $+$ $\measuredangle$ $ACO$ $+$ $\measuredangle$ $COA=0$

After this we can write $2\times \measuredangle$ $OAC$ $+$ $\measuredangle$ $COA=0$
and resubstitute $\measuredangle$ $COA$ as $2\times \measuredangle$ $CBA$
We get, $2\times \measuredangle$ $OAC$ $+$ $2\times \measuredangle$ $CBA=0^\circ (\text{mod}\ 180^\circ)$
which is equivalent to writing as $2\times \measuredangle$ $OAC$ $+$ $2\times \measuredangle$ $CBA=180^\circ (\text{mod}\ 180^\circ)$
Divide both sides by $2$, and proceed to get $\measuredangle$ $OAC$ + $\measuredangle$ $CBA$ = $90^\circ \ (\text{mod}\ 90^\circ)$
using : If a ≡ b (mod c) and gcd(c, d) = g then a/d ≡ b/d (mod c/g)
Hence proved.
$\angle CBA \ne \angle CXA$, but $\angle CBA +\angle CXA =\pi$.