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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 $\phi$.

Sketch of the problem

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.

Sketch of the solution

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.

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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.

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  • $\begingroup$ How do I know that $EHC$ is a straight line and that $H$ belongs to that line?. I don't see very clearly how do you conclude $\angle BEC = 15^{\circ}$, perhaps can you help me with this?. From there on I became confused on how to calculate $EP$, $PC$ and $PB$ as you mentioned. I can only guess that with those sides known using cosines law I can calculate angle $\angle PBC$ but maybe can you add more details please? $\endgroup$ – Chris Steinbeck Bell Mar 31 '19 at 5:35
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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 $\angle ABC = 45^\circ$.

enter image description here

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  • $\begingroup$ How does symmetry proves that the triangle is isosceles?. Maybe can you add some drawing so I could spot what were you referring?. I must say that in my drawing $D \notin AB$ but rather $D \in BH$. Hence how can I prove $DC \cong DB$?. Does it exist a way to do that?. But I must say that the part where you established $FC \perp AB$ makes easy to find $\angle ABC = 45 ^{\circ}$. I'd hope you please help me to answer those unattended questions. :) $\endgroup$ – Chris Steinbeck Bell Mar 31 '19 at 5:44
  • $\begingroup$ @ChrisSteinbeckBell I've added an image. $\endgroup$ – Michael Biro Mar 31 '19 at 13:00

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