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The problem is as follows:

In a regular hexagon $ABCDEF$ a point $G$ is located on $CD$ such that $AG$ and $BE$ intersect at point $R$. If $\angle AGD = 110^\circ$. Find $\angle BCR$.

The alternatives given in my book are:

$\begin{array}{ll} 1.&40^\circ\\ 2.&35^\circ\\ 3.&50^\circ\\ 4.&45^\circ\\ \end{array}$

I'm stuck on this problem. However since this problem does not include a graph. I did my best effort as an interpretation of what has been mentioned.

And I believe the problem is indicating this figure:

Sketch of the problem

But what to do from here?. I cannot spot which sort of theorem or identity relying on euclidean geometry approach could be used to solve this problem?.

Can someone help me here?. I've attempted to look for congruence or maybe construction but I just can't spot else. Please include a drawing in your answer. Please help.

Chris Steinbeck Bell
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    Why is it that you've accepted almost-no answers to your questions over the years? – Blue Jan 13 '21 at 03:46
  • @Blue I'm sorry for that. I'm doing my best to keep up to date, it is not easy to review and understand all the answers. I have to do that to take a good judgement and also my health isn't very good so I cannot use internet connection too often. But I'll take note to improve that. – Chris Steinbeck Bell Jan 13 '21 at 23:37

1 Answers1

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The answer is 50 degrees.

First, construct a straight line from C to R. Then, label the point at which this meets AF as $G'$.

Recall that in a regular hexagon, the internal angles are $120^\circ$. This implies that $\angle AFE=120^\circ$, and $\angle BEF=60^\circ$.

Also, note that the lines CG' and AG are "reflections" of each other across the line BE. Thus, $\angle CG'F=\angle AGD=110^\circ$ by symmetry. Note that G'REF is a quadrilateral, which has an internal angle sum of $360 ^\circ$. Hence, we conclude that $\angle G'RE=360^\circ-120^\circ-60^\circ=70^\circ$. This of course means that $\angle BRC=\angle G'RE=70^\circ$, due to vertically opposite angles being equal. Note that $\angle RBC =60 ^\circ$. From these, we can conclude that $\angle BCR=50^\circ$ since the angles of the triangle BCR add up to $180 ^\circ$.

Gerry T.
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  • I think you need to correct $\angle RG'F =360-(120+110+60)=70^\circ$ and also note that $\angle G'RE=\angle BRC$ then $\angle RBC=60^\circ$ ,therefore in $\triangle CBR$ you use the identity that all angles add up to $180^\circ$ and because of such $\angle BCR=50^\circ$ for that the rest I think its okay. Would you mind to include these corrections? as part of your answer?. – Chris Steinbeck Bell Jan 16 '21 at 03:54
  • @ChrisSteinbeckBell You're right, it seems that I have made a typo where I accidentally wrote $\angle RG'F=360-....=70 ^\circ$ when it should've been $\angle G'RE$ instead. – Gerry T. Jan 16 '21 at 06:47
  • I will also add some more lines for further clarification, thanks! – Gerry T. Jan 16 '21 at 06:47