10.2: Cross-ratio

The following theorem gives some quantities expressed in distances or angles that do not change after inversion.

Theorem $\PageIndex{1}$

Let $ABCD$ and $A'B'C'D'$ be two quadrangles such that the points $A',B',C'$, and $D'$ are the inverses of $A,B,C$, and $D$ respectively.

Then

(a)

$\dfrac{AB \cdot CD}{BC \cdot DA} = \dfrac{A'B' \cdot C'D'}{B'C' \cdot D'A'}$.

(b)

$\measuredangle ABC + \measuredangle CDA \equiv -(\measuredangle A'B'C' + \measuredangle C'D'A')$.

(c) If the quadrangle $ABCD$ is inscribed, then so is $\square A'B'C'D'$.

Proof

(a). Let $O$ be the center of the inversion. According to Lemma 10.1.1, $\triangle AOB \sim \triangle B'OA'$. Therefore,

$\dfrac{AB}{A'B'} = \dfrac{OA}{OB'}.$

Analogously,

$\dfrac{BC}{B'C'} = \dfrac{OC}{OB'}$, $\dfrac{CD}{C'D'} = \dfrac{OC}{OD'}$, $\dfrac{DA}{D'A'} = \dfrac{OA}{OD'}$.

Therefore,

$\dfrac{AB}{A'B'} \cdot \dfrac{B'C'}{BC} \cdot \dfrac{CD}{C'D'} \cdot \dfrac{D'A'}{DA} = \dfrac{OA}{OB'} \cdot \dfrac{OB'}{OC} \cdot \dfrac{OC}{OD'} \cdot \dfrac{OD'}{OA}.$

Hence (a) follows.

(b). According to Lemma 10.1.1,

$$\begin{array} {l} {\measuredangle ABO \equiv -\measuredangle B'A'O, \measuredangle OBC \equiv -\measuredangle OC'B',} \\ {\measuredangle CDO \equiv -\measuredangle D'C'O, \measuredangle ODA \equiv -\measuredangle OA'D'.} \end{array}$$

By Axiom IIIb,

$\measuredangle ABC \equiv \measuredangle ABO + \measuredangle OBC$, $\measuredangle D'C'B' \equiv \measuredangle D'C'O + \measuredangle OC'B'$,
$\measuredangle CDA \equiv \measuredangle CDO + \measuredangle ODA$, $\measuredangle B'A'D' \equiv \measuredangle B'A'O + \measuredangle OA'D'$,

Therefore, summing the four identities in 10.2.1, we get that

$\measuredangle ABC +\measuredangle CDA \equiv -(\measuredangle D'C'B' + \measuredangle B'A'D')$.

Applying Axiom IIIb and Exercise 7.4.5, we get that

$\begin{array} {rcl} {\measuredangle A'B'C' + \measuredangle C'D'A'} & \equiv & {-(\measuredangle B'C'D' + \measuredangle D'A'B') \equiv} \\ {} & \equiv & {\measuredangle D'C'B' + \measuredangle B'A'D'.} \end{array}$

Hence (b) follows.

(c). Follows from (b) and Corollary 9.3.2.