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Hexagon $ABCDEF$ has sides $AB$ and $DE$ of length $2$, sides $BC$ and $EF$ of length $7$, and sides $CD$ and $AF$ of length $11$, and it is inscribed in a circle. Compute the diameter of the circle.

According to Ptolemy's Theorem, \begin{equation*} 2(11) + 7 \left\vert \overline{\mathit{AD}} \right\vert = \left\vert \overline{\mathit{AC}} \right\vert \left\vert \overline{\mathit{BD}} \right\vert . \end{equation*} I am told that $\overline{\mathit{AD}}$ is a diameter of the circle, in which case, $\triangle{ACD}$ and $\triangle{ABD}$ are right triangles. By Pythagorean Theorem, \begin{equation*} \left\vert \overline{\mathit{AC}} \right\vert^{2} = \left\vert \overline{\mathit{AD}} \right\vert^{2} - 11^{2} \qquad \text{and} \qquad \left\vert \overline{\mathit{BD}} \right\vert^{2} = \left\vert \overline{\mathit{AD}} \right\vert^{2} - 2^{2} . \end{equation*} The length of $\overline{\mathit{AD}}$ can be computed to be $14$.

Here is my question: Why is $\overline{\mathit{AD}}$ the diameter of the circle?

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  • $\begingroup$ This is all overkill. The hexagon's rotational symmetry follows immediately from it being an inscribed polygon with cyclic edge lengths. $\endgroup$ – user334732 Mar 29 '18 at 7:39
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According to the Inscribed-Angle Theorem, $\angle\mathit{ABC} \cong \angle\mathit{DEF}$, and, according \ to the Side-Angle-Side Theorem, $\triangle\mathit{ABC} \cong \triangle\mathit{DEF}$. In particular, $\angle\mathit{BAC}$ \ $\cong \angle\mathit{EDF}$, and $\overline{\mathit{AC}} \cong \overline{\mathit{DF}}$. Again, according to the Inscribed-Angle Theorem, $\angle\mathit{ADC} \cong \angle\mathit{DAF}$.

Likewise, $\angle\mathit{BAF} \cong \angle\mathit{CDE}$, and $\overline{\mathit{BF}} \cong \overline{\mathit{CE}}$.

Since \begin{equation*} \mathrm{m}\angle\mathit{BAC} + \mathrm{m}\angle\mathit{CAD} + \mathrm{m}\angle\mathit{DAF} = \mathrm{m}\angle\mathit{BAF} = \mathrm{m}\angle\mathit{CDE} = \mathrm{m}\angle\mathit{EDF} + \mathrm{m}\angle\mathit{FDA} + \mathrm{m}\angle\mathit{ADC} , \end{equation*} $\mathrm{m}\angle\mathit{CAD} = \mathrm{m}\angle\mathit{FDA}$.

According to the Angle-Side-Angle Theorem, $\triangle\mathit{ADF} \cong \triangle\mathit{DAC}$; in particular, $\angle\mathit{AFD} \cong \angle\mathit{DCA}$. These angles are supplementary, though. Consequently, they are right angles. According to the Inscribed-Angle Theorem, $\overline{\mathit{AD}}$ is a diameter of the circle.

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Opposite triangle is congruent. Then that makes a rectangle. enter image description here

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  • $\begingroup$ I appreciate your comment and diagram showing that three pairs of congruent triangles are inscribed in the circle. The Inscribed-Angle Theorem and the Alternating-Angle Theorem have to be cited to conclude that the pair of unmarked sides and the pair of sides marked with "|" are parallel. $\endgroup$ – user74973 Mar 19 '18 at 18:36
  • $\begingroup$ @user74973 ○ and × are equal since both are angle of equal two side. $\endgroup$ – Takahiro Waki Mar 19 '18 at 18:49
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    $\begingroup$ @user74973 Very agree. $\endgroup$ – Takahiro Waki Mar 19 '18 at 19:16
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    $\begingroup$ Your argument doesn't show that you have a rectangle; it shows that you have a parallelogram. $\endgroup$ – user74973 Mar 26 '18 at 2:58
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    $\begingroup$ Yep. The only parallelograms that can be inscribed in a circle are rectangles. $\endgroup$ – user232552 Mar 26 '18 at 2:59
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Note that the measure of an inscribed angle is half the measure of the arc which is covered by the angle.

Thus, if two angles share the same arc, they have the same measure.

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  • $\begingroup$ @user An arc subtends an angle at centre, its half is subtended at any point on circumference of circle. This can be proved by drawing a radius joining the centre and that point. But this is the converse. $\endgroup$ – King Tut Mar 15 '18 at 14:19
  • $\begingroup$ Oh it was not clear for me, sorry! $\endgroup$ – King Tut Mar 16 '18 at 2:00
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The lengths of the edges $\overline{ABCDEFA}$ go $2,7,11,2,7,11$ which is cyclic of order $2$, so the inscribed hexagon has rotational symmetry if turned through $\frac{2\pi}{2}$. It follows immediately that any diagonal drawn between pairs of opposite vertices is a perfect diameter.

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