Quadrilateral in Complex Plane is Cyclic iff Cross Ratio of Vertices is Real
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Theorem
Let $z_1, z_2, z_3, z_4$ be distinct complex numbers.
Then:
- $z_1, z_2, z_3, z_4$ define the vertices of a cyclic quadrilateral
if and only if their cross-ratio:
- $\paren {z_1, z_3; z_2, z_4} = \dfrac {\paren {z_1 - z_2} \paren {z_3 - z_4} } {\paren {z_1 - z_4} \paren {z_3 - z_2} }$
is wholly real.
Proof
Let $z_1 z_2 z_3 z_4$ be a cyclic quadrilateral.
By Geometrical Interpretation of Complex Subtraction, the four sides of $z_1 z_2 z_3 z_4$ can be defined as $z_1 - z_2$, $z_3 - z_2$, $z_3 - z_4$ and $z_1 - z_4$.
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, the opposite angles of $z_1 z_2 z_3 z_4$ sum to $\pi$ radians.
By Complex Multiplication as Geometrical Transformation, it follows that:
\(\ds \angle \, z_1\) | \(=\) | \(\ds \map \arg {\dfrac {z_1 - z_2} {z_1 - z_4} }\) | ||||||||||||
\(\ds \angle \, z_2\) | \(=\) | \(\ds \map \arg {\dfrac {z_3 - z_2} {z_1 - z_2} }\) | ||||||||||||
\(\ds \angle \, z_3\) | \(=\) | \(\ds \map \arg {\dfrac {z_3 - z_4} {z_3 - z_2} }\) | ||||||||||||
\(\ds \angle \, z_4\) | \(=\) | \(\ds \map \arg {\dfrac {z_1 - z_4} {z_3 - z_4} }\) |
Thus:
\(\ds \angle \, z_1 + \angle \, z_3\) | \(=\) | \(\ds \pi\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \arg {\dfrac {z_1 - z_2} {z_1 - z_4} } + \arg \paren {\dfrac {z_3 - z_4} {z_3 - z_2} }\) | \(=\) | \(\ds \pi\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \arg {\paren {\dfrac {z_1 - z_2} {z_1 - z_4} } \paren {\dfrac {z_3 - z_4} {z_3 - z_2} } }\) | \(=\) | \(\ds \pi\) | Argument of Product equals Sum of Arguments | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \Im {\dfrac {\paren {z_1 - z_2} \paren {z_3 - z_4} } {\paren {z_1 - z_4} \paren {z_3 - z_2} } }\) | \(=\) | \(\ds 0\) | Argument of Negative Real Number is Pi |
Hence the result.
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Exercise $6$.