Without a calculator, determine whether chords $AB, AC \text{ and }BD$ can divide a circle into five equal-area regions. Without a calculator, determine whether chords $AB, AC \text{ and }BD$ can divide a circle into five equal-area regions.

Context: I just made up this question.
I have only been able to answer the question with a calculator. Surely there must be some clever way to answer the question without a calculator.
Answer that uses a calculator:
Proof by contradiction: assume the answer is yes.
Let radius $=1$.
Call the centre $O$.
$\alpha=\angle{AOB}$
$\beta=\angle{AOC}$
$\text{Area}_{\text{minor segment }AB}=\dfrac12(\alpha-\sin{\alpha})=\dfrac{\pi}{5}\implies \alpha\approx2.1131$
$\text{Area}_{\text{segment }ADC}=\dfrac12(\beta-\sin{\beta})=\dfrac{2\pi}{5}\implies \beta\approx2.8248$
$\text{Area}_{\text{triangle}}=\dfrac{\pi}{5}\text{ and }AB=2\sin{\dfrac{\alpha}{2}}\implies ...\implies \angle{BAC}=\arctan{\left(\dfrac{\pi}{5\sin^2{\dfrac{\alpha}{2}}}\right)}\approx{0.692}$
But $\angle{BAC}=\dfrac12{\angle{BOC}}=\dfrac12{\left(2\pi-\alpha-\beta\right)}\approx{0.673}$, contradiction.
Therefore the answer is no.
UPDATE: I found essentially the same question, except it does not request a calculator-free solution. So it doesn't answer my question.
 A: Second try (cleaner) Please refer to the diagram in my other answer.
By naming $2\theta=\widehat{BOA},\, 2\varphi=\widehat{AOC}$ and by imposing that the circle segments on $AB,AC,BD$ have the correct areas we end up with
$$ \theta-\frac{1}{2}\sin(2\theta)=\frac{\pi}{5},\qquad \varphi-\frac{1}{2}\sin(2\varphi)=\frac{2\pi}{5}.\tag{A}$$
In the triangle $ABE$ we have $AB=2\sin\theta$ and $\widehat{ABE}=\pi-(\theta+\varphi)$, so
$$ [ABE] = -\sin^2(\theta)\tan(\theta+\varphi) \tag{B}$$
and given $(A)$ it is enough to show that $(B)$ cannot be equal to $\frac{\pi}{5}$. 
$f(x)=x-\frac{\sin(2x)}{2}$ is non-negative, convex and increasing from $0$ to $\pi/2$ on $[0,\pi/2]$, with $f'(x)=2\sin^2(x)$.
We have $\theta\approx \pi/3$ and by a single step of Newton's method with starting point $\pi/2$ we have $\varphi\approx 9\pi/20$.
Precisely $\theta > \pi/3$ and $\varphi < 9\pi/20$, and the error of these approximations is controlled by the fact that $1\leq f'\leq 2$ in $[\pi/4,\pi/2]$. In particular
$$ \theta < \frac{\pi}{3}+\left(\frac{\pi}{5}-f(\pi/3)\right) $$
$$ \varphi > \frac{9\pi}{20}-\left(f(9\pi/20)-\frac{2\pi}{5}\right)\tag{C}$$
such that $[ABE]$ is very close to $\frac{3}{4}\tan\left(\frac{13}{60}\pi\right)$. The impossibility of the pentasection then follows from the proof of the following inequality:
$$ \tan\left(\frac{13\pi}{60}\right)<\frac{4\pi}{15}=\frac{16\pi}{60}\tag{D} $$which is equivalent to
$$ \frac{1-\tan(\pi/30)}{1+\tan(\pi/30)} < \frac{4\pi}{15}\tag{E} $$
or to
$$ \tan\left(\frac{\pi}{30}\right) > \frac{15-4\pi}{15+4\pi} \tag{F}$$
which just follows from
$$ \tan\left(\frac{\pi}{30}\right) > {\color{red}{\frac{\pi}{30}}} > \frac{15-4\pi}{15+4\pi} \tag{G}$$
since $\pi > \frac{15}{8}(\sqrt{113}-9).$
A: 
Let us assume that the radius of the circle is $1$ and let $\widehat{BOA}=2\theta$.
Then $\theta$ is the unique solution of $\theta-\frac{1}{2}\sin(2\theta)=\frac{\pi}{5}$ and $AB=\sin(2\theta)$.
$[AEB]=\frac{\pi}{5}$ then implies $ME=\frac{\pi}{5\sin\theta}$ and $AE=EB=\sqrt{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}$.
The height of the circle segment with base $AB$ is $1-\cos\theta$, so by naming $P$ the antipode of $N$ we have $EP=1-\cos\theta+\frac{\pi}{5\sin\theta}$ and $EN=1+\cos\theta-\frac{\pi}{5\sin\theta}$. By the chords theorem $EN\cdot EP = EC\cdot EA$, hence
$$ EC = \frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sqrt{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}}$$
and by the similarity between $EAB$ and $ECD$ we have
$$ CD = \frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot 2\sin\theta$$
and
$$ \widehat{DOC} = 2\arcsin\left(\frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot \sin\theta\right).$$
The length of $OE$ is $\frac{\pi}{5\sin\theta}-\cos\theta$, hence the area of the quadrilateral $DOCE$ equals
$$ [DOCE] = \left(\frac{\pi}{5\sin\theta}-\cos\theta\right)\left(\frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot \sin\theta\right) $$
and the area of the portion of the circle bounded by $E,D,C$ equals
$$\arcsin\left(\frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot \sin\theta\right)-\left(\frac{\pi}{5\sin\theta}-\cos\theta\right)\left(\frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot \sin\theta\right).$$
Few steps of Newton's method give $\theta\approx 1.05657$, hence the last expression is $\approx 0.552681$ while $\pi/5\approx 0.628319$, proving that the wanted pentasection is impossible.
