Directed angles problem: For $A$, $B$, $C$ on $\bigcirc O$, show that $\angle OAC=90^\circ-\angle CBA$

Question

Directed angles problem from Evan Chen's book:

Points $A$, $B$, $C$ lie on a circle with center $O$. Show that $$\measuredangle OAC=90^\circ-\measuredangle CBA$$ (This is not completely trivial.) Hints: 8 530 109

(original problem image)

Hello Stackexchange! I can't solve this problem because I didn't understand directed angles completely. Solution with deep explanation would be amazing!

Answer 1

So we start by determening the value of $\measuredangle COA$. Since O is a central angle and $\measuredangle CBA$ is an inscribed angle based on the same segment $\overline{CA}$. Since the central angle is two times the according inscribed angle (look here for more information) we get $\measuredangle COA = 2\cdot\measuredangle CBA = 2\beta$. Furthermore, looking at triangle $\triangle COA$, we notice that $\overline{CO}=\overline{AO}=R$, hence, the angles opposite to the sides will also be equal $\Rightarrow \measuredangle ACO = \measuredangle CAO = x$. Now we have an easy equation to solve by adding up all the angles: $$x+x+(2\beta) = 180° = 2x+2\beta$$ $$x+\beta =90°$$ $$x = 90-\beta$$ If you have any questions, just ask them in comments!

Answer 2

Extend the line $OA$ so that it intersects the circle at two points. One of them is $A$; the other one we can call it $A'$.

Below is one possible diagram; it is also possible that $O$ is not inside the triangle $\Delta ABC$, but the relationship between angles I mentioned below still holds.

I don't know how theorems are named in Evan's book, but all things I call theorem below should exist in his book.

One method:

One theorem says the arc corresponding to the inscribed angle $\angle OAC$ and $\angle CBA$ are arc $A'C$ and arc $AC$ respectively, and they together form half a circle.

The angle with vertex is at $O$ [called the central angle] corresponding of one full circle is $360$ degrees (like if you start at the ray $OA$ and swing a full round) whereas the central angle corresponding to half a circle is half of $360$ degrees. For the same arc the inscribed angle is always half the central angle. So inscribed angles corresponding to half a circle is half of half of $360$ degrees, which is $90$ degrees.

Since $\angle OAC$ and $\angle CBA$ together correspond to arcs adding up to half a circle, they adds up to $90$ degrees.

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Justification the adds-up process above:

A theorem says all inscribed angle of the same arc equal each other. So $\angle OAC = \angle A'BC$. Thus $$\angle OAC+ \angle CBA = \angle A'BC + \angle CBA = \angle A'BA$$ The angle $\angle A'BA$ is an inscribed angle corresponding to the diameter, so it is $90^{\circ}$.

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Same theorems applied in another order:

One theorem says, for the arc $A'C$, the central angle $\angle A'OC$ is twice the inscribed angle $\angle OAC$. For the arc $AC$ the central angle $\angle COA$ is also twice the inscribed angle $\angle CBA$.

In other words, $$\angle A'OC = 2\angle OAC, \ \ \ \angle COA = 2 \angle CBA.$$

We have $\angle A'OC + \angle COA = \angle AOA' = 180^{\circ}$. Substituting in the two equations above and dividing both sides by 2, we get $$\angle OAC + \angle CBA = 90^{\circ}.$$ Rearrange for the desired equality.