Example of a hexagon inscribed in a circle

Question

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?

Answer 1

Opposite triangle is congruent. Then that makes a rectangle.

Answer 2

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.

Answer 3

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.

Answer 4

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.