Page-15, Lemma 1.30: Points $A,B,C$ lie on a circle with center $O$. Show that $\measuredangle OAC= 90 – \measuredangle CBA$
Here we $\measuredangle CBA$ means the directed angle. I've seen this post and linked already but I think neither has solutions in the way which was intended by the hint for the question.
I think so I got the idea how to solve it, but I'd like my solution be checked because this is one of my first times working with directed angles. Here goes:
Proof:
Begin by constructing the reflection of $A$ about $O$, call it $A'$. Notice that $\triangle ACA'$ is right angled at $C$, hence:
$$ 90 +\measuredangle A_o A C + \measuredangle CA_o A=0$$
And by the inscribed angle theorem, I note that $ \measuredangle CBA = \measuredangle CA_o A$ and also $\measuredangle A_o A C= \measuredangle OAC$, upon substituting, I get:
$$ \measuredangle OAC + \measuredangle CBA = 90$$
Is my solution well written? How could I make it better?
Here is a drawing I used for 'thinking' about the set up(slight error , reflected should be $A'$ not $A_o$):
And for $B$ on other side:
Turns out that exact same argument as said for the other case applies here (if the first proof was valid).
Note: Acc. to Evan chen's book sum of directed angle is zero,
page-12
Best Answer
Here is an alternative approach -
Say $\angle ABC = \theta \ $. By inscribed angle theorem,
$\angle AOC = 2 \angle ABC = 2 \theta$.
As $OA = OC$, $\triangle AOC$ is isosceles triangle. So,
$\angle OAC = \angle OCA = \dfrac{180^0-2\theta}{2} = 90^0 - \theta$