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Lines $ CF,GE $ are parallel to diagonal $DB$.

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Are the cyclic quadrilaterals $ CBGDC,FBGDF $ (with same sides and angles) congruent? If not, how are they be described?

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  • $\begingroup$ It would be clearer to spell out the four cyclic quadrilaterals you are concerned with. Clearly $CBGD$ and $FDEB$ are congruent (by reflection), but $CBGD$ and $FBGD$ need not be (assuming distinctness of $C$ and $F$ and of $G$ and $E$, as the illustration indicates). $\endgroup$ – hardmath Sep 19 '16 at 6:49
  • $\begingroup$ Changed to include only two, since the others are anyhow symmetric. $\endgroup$ – Narasimham Sep 19 '16 at 7:36
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Yes. Draw the line $l$ through the center $O$ perpendicular to $DB$. Then $l$ passes through the midpoint of $DB$. Moreover $l$ is orthogonal to $CF$ and $GE$ because the two latter chords are parallel to parallel to $DB$. Therefore $l$ passes through the midpoints of $CF$ and $GE$. Consequently a reflection in $l$ maps one quadrilateral into the other.

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  • $\begingroup$ Note that this is the symmetry I called attention to in my Comment, and which led to a clarification of the Question that obviates its consideration. The reflection pointed out does not produce congruence of $CBGDC,FBGDF$ because the "lower" triangles are not swapped. $\endgroup$ – hardmath Sep 20 '16 at 0:05
  • $\begingroup$ I can prove only things that are true. The ones that are not... I do not prove :)... And I did not have time to read your comment... $\endgroup$ – Futurologist Sep 20 '16 at 2:20

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