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# 10.2: Cross-ratio

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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.
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