How to find the angle when an hexagon is rotated along one of its corners?

Question

I've been going in circles with this problem but I've found some solution from looking it many times:

The problem is as follows:

A certain protein is under investigation in a laboratory in Taichung. The atoms are arranged along the corners of an hexagon and to examine its optic properties, the crystal is rotated counterclockwise $30^{\circ}$ so that the opposing side forms a $90^{\circ}$ as shown in the figure. If it is known that the light passing through the crystal bends exactly the angle $\angle ABC$. Find the bending angle labeled as $\phi$.

The alternatives shown in my book are as follows:

$\begin{array}{ll} 1.&60^{\circ}\\ 2.&37^{\circ}\\ 3.&53^{\circ}\\ 4.&75^{\circ}\\ 5.&45^{\circ}\\ \end{array}$

What I did in my attempt to solve the problem is sumarized in the sketch from below.

In other words, I did spotted that there is an hexagon in $EAHDGF$ so that the total sum of its interior angles would be equal to $6\times 120^{\circ}$. Although some of its corners have different angles. Since what it is being asked is $\angle ABC$. It's already known that $\angle ABH = 30^{\circ}$ as $\angle AEB= 60^{\circ}$.

From this I inferred the following:

$6\left( 120^{\circ}\right)=90^{\circ}+4\left(120^{\circ}\right)+180^{\circ}-2\omega$

Therefore:

$2\omega=180^{\circ}-2\left(120^{\circ}\right)+90^{\circ}$

$2\omega=270^{\circ}-240^{\circ}=30^{\circ}$

$\omega=15^{\circ}$

Now all that is left to do is to sum $\omega +30^{\circ}=\phi$

therefore:

$15^{\circ}+30^{\circ}=45^{\circ}$

However to establish this answer I had to take for granted that $\triangle BDC$ is isosceles. This part is where I'm still stuck as I couldn't find a way to prove that. Can somebody help me with this matter?. I'd like to know if there are other ways to get this answer. By looking in my book the answer I got is correct. But still I feel dubious if what I did was the right thing to do.

Answer 1

Draw EHC, a line. Angle BEC is 15 degrees.
As length of EB and EC are know, BC can be calculated.
Let P be the point of intersection of EC and AB.
Using triangle EAP, calculate EP, PC, PB and angle PBC.

Answer 2

You can note that the triangle is isosceles by symmetry. Since each hexagon is a $30^\circ$ rotation of the other, $\angle ABC = \angle FCB$, so triangle $BCD$ is isosceles. Furthermore, $FC$ is perpendicular to $AB$, so they form a right triangle as well. Therefore $\angle ABC = 45^\circ$.

Edited: Here is an image of what I mean. Note the symmetry that means that $LB = LC$. Since $AB$ and $CF$ are perpendicular, that means $BCL$ is an isosceles right triangle, so $\angle ABC = 45^\circ$.