This question was from a math competition, and I can't seem to solve it. Please refer to the picture of the problem I'm trying to solve.
Attempt #1
Draw a line from B to E and from A to F and lable the point where the line AF intersects BH, P. Then from the symmetry of the picture, $AB$ and $EF$ are equal, and since $AE$ is a diameter, $\angle AFE$ and $\angle ABE$ are right. Because triangles $\triangle AFE$ and $\triangle ABE$ share a side, have two sides the same length and both have a right angle, they are congruent $\triangle AFE \cong \triangle ABE$, so $\angle BAF$ is right. From there it is straightforward to find that $\triangle ABP \sim \triangle ABE$ due to their angles.
Let $r$ be the radius of the circle. Since $BP=\frac{8}{\cos(20)}$ and $AP=8\tan(20)$, we can write (by similar triangles) that $$\frac{AE}{AB}=\frac{BP}{AP} \Rightarrow \frac{2r}{8}=\frac{\frac{8}{\cos(20)}}{8\tan(20)}$$
From this we find that $r=\frac{4}{\sin(20)}$, which will obviously not yield one of the answers for the area of the circle. I'm not 100% sure where I went wrong but I think it might be somewhere assuming $\angle BAF=90^\circ$ but at the same time I'm not really sure.
Attempt #2
This 'solution' is somewhat stupid but should work because it's multiple choice. Since AE is a diameter then $\angle ABE$ is right so applying the pythagorean theorem gives $$8^2+(BE)^2=(2r)^2 \Rightarrow 4^2 + \bigg(\frac{BE}{2} \bigg)^2 =r^2$$ From the answers $r$ is an integer and the only pythagorean triple that would work here is $(3,4,5)$ so $r=5$ but again this doesn't give any of the answers.
Is the question just simply wrong, or am I wrong in my logic?
From all the parallel lines, consider the angles at circumference:
$$\begin{align*} \angle BHC &= \angle CGD = \angle DFE = 20^\circ\\ \angle BAE &= \angle BAC + \angle CAD + \angle DAE\\ &= \angle BHC + \angle CGD + \angle DFE\\ &= 60^\circ\\ \end{align*}$$
Consider right-angled triangle $\triangle ABE$,
$$\begin{align*} \cos \angle BAE &= \frac{AB}{AE}\\ AE &= 2AB\\ \frac{AE}2 &= AB = 8\text{ cm} \end{align*}$$