A circle is divided into $5$ parts as shown in the diagram and parts are colored either red or green. Find which area is bigger. In the given diagram, there are $5$ points $A, B, C, D$ and $E$ on the circumference of the circle such that $\angle ABC = \angle BCD = \angle CDE = 45^{\circ}$ and $O$ is the center of the circle.
Sectors made by $AB$ and $DE$, and area of the circle between $BC$ and $CD$ are highlighted in green. Area of the circle between $AB$ and $BC$, and between $CD$ and $DE$ are highlighted in red.
Which area is bigger, the area highlighted in red or the area highlighted in green?

This was sent to me by someone. While I solved the problem (given below), the sender said that the source solution arrived at the conclusion that points $A$, $O$ and $E$ are collinear and $OC \perp AE$, so $\displaystyle \angle OCB = \angle OCD = \frac{45^{\circ}}{2}=22.5^{\circ}$. While I agree with points being collinear and $OC \perp AE$ but that cannot obviously be the reason for the angles being equal. In fact the solution does not depend on them being equal as we can see. I am seeking help in establishing $\angle OCB = \angle OCD$ if that is indeed true, which I cannot see how one can conclude based on what is given.
My solution: Say, $\angle OCB = \theta$. Then, $\angle ACB = \angle OCD = (45^{\circ}-\theta)$ and $\angle DCE = \theta$.
Segment $AB= \displaystyle r^2 \left[\frac{\pi}{4}-\theta-\sin(45^{\circ}-\theta)\cos(45^{\circ}-\theta)\right]$
Segment $DE= \displaystyle r^2 \left[\theta-\sin \theta \cos \theta\right]$
$\triangle OBC = r^2 \sin \theta \cos \theta$
$\triangle ODC = r^2 \sin(45^{\circ}-\theta)\cos(45^{\circ}-\theta)$
Section $BOD = \dfrac{\pi}{4} r^2$
Adding all of the above, total area in green $= \dfrac{\pi}{2} r^2$. So the red area has to be the same too.
In addition to my question on $OC$ being bisector of $\angle BCD$, let me also know if any of you have a simpler solution.
 A: Reflect the area $ABC$ such that $A\to A'$ and $B\leftrightarrow C$, and similarly reflect the area $CDE$ such that $C\leftrightarrow D$ and $E\to E'$. All sectional parts then have a vertex at $C$. We now have a limiting case of the Pizza Theorem ("limiting" because the center of the cutter is on the circumference). This theorem says that the white and the grey parts in the figure have the same area.

A: For simplicity say $r=1$.
$$Green_1 = {\pi x\over 360} - {\sin x \over 2}$$
$$Green_2={\sin (x+90) \over 2} + {\sin(180- x) \over 2} +  {\pi \over 4}$$
$$Green_3 = {\pi (90-x)\over 360} - {\sin (90- x) \over 2}$$
All the $\sin $ cancels so we have $$ Green = {\pi \over 2} = Red$$

A: Not an answer by any stretch of the imagination, I just want to illustrate the dangers of drawing pictures that are too "nice" or too symmetrical. We could just as well have the following configuration, so there's clearly no bisector.

A: No, those angles don't have to be equal. Angle subtended by chord being half the angle at the centre gives that $AOC=COE=90^\circ$. Assume this, and let $C'$ be the point opposite $C$, so that $ACEC'$ is a square. Now if $B$ is placed anywhere on the circumference between $A$ and $C'$, the angle $ABC$ will automatically be $45^\circ$. There is a line through $C$ at $45^\circ$ to $BC$, which meets the circle between $C'$ and $E$; this is a valid position for $D$ such that all three angles are $45^\circ$. So $OCB$ can be anywhere between $0^\circ$ and $45^\circ$.
