Throughtout, $\Re$ denotes cross-ratio (I couldn’t get \Re to work in mathjax).

4.6

$\require{textcomp}$ We simply apply the cocycle identity of cross-ratio: \(\begin{aligned} \Re(A,B;E,C)&=\Re(A,B;D,C)\Re(A,B;E,D)\\ &=-\Re(A,B;E,D)\\ &=-\Re(A',C';B',D)&\text{via projection from }O'\\ &=-\Re(A,C;B,D)&\text{via projection from }O\\ &=-(1-\Re(A,C;B,D))&\text{by Lemma }\ref{permutecrossratio}\\ &=-2.\end{aligned}\)

4.8

First, $\mathrm{ratio}(A,H,C)=1/4$, so if $Q$ is the point at infinity on $AC$, then $\Re(A,H;C,Q)=1/4$. Hence $\Re(Q,H;A,C)=4/3$. Choose $n$ on $C$ such that $d(n \cap PH,n \cap PA)=d(n \cap PH, C)$. (That is, take the line $g$ parallel to $PA$ through $C$, and then take the parallel $h$ to $PC$ through $PH\cap g$. The point $n$ is the line $CX$ where $X=h\cap PA$). Then \(\Re(n \cap \ell',n \cap PH;n \cap PA,C)=4/3,\) so $d(n \cap \ell’,n \cap PH)=d(n \cap PA, C)$.

4.9

(i) As usual, we set two reference points $O$ and $I$ on $\ell$ so that we can write each point in coordinate form; write $(x,1)$ for the point $xI+O$. We will assume $I$ is the point at infinity on $\ell$, so that it is not one of our points. Let \(A=(a,1), \quad B=(b,1), \quad C=(c,1),\quad D=(d,1).\) In particular, we can now write expressions such as $d(B,A)=b-a$ (n.b., we are using ‘directed distance’ here, but it will not matter in what follows). Let $P$ be the point $(p,1)$. Then for each $X$ in ${C,D}$, we have \(\begin{aligned} \Re(X,B;P,A)&=\frac{(p-x)(a-b)}{(p-b)(a-x)}\end{aligned}\) and hence \(\begin{aligned} \Re(C,B;P,A)+\Re(D,B;P,A)&=\frac{(p-c)(a-b)}{(p-b)(a-c)}+\frac{(p-d)(a-b)}{(p-b)(a-d)}\\ &=\frac{a-b}{p-b}\left( \frac{p-c}{a-c}+\frac{p-d}{a-d}\right)\\ &=\frac{d(A,B)}{d(P,B)}\left( \frac{d(P,C)}{d(A,C)}+\frac{d(P,D)}{d(A,D)}\right)\end{aligned}\) Now $d(P,C)=d(P,A)+d(A,C)$, and $d(P,D)=d(P,A)+d(A,D)$, and so \(\begin{aligned} \Re(C,B;P,A)+\Re(D,B;P,A) &=\frac{d(A,B)}{d(P,B)}\left( \frac{d(P,A)}{d(A,C)}+\frac{d(P,A)}{d(A,D)}+2\right)\\ &=\frac{d(P,A)}{d(P,B)}\left( d(A,B)\cdot \left(\frac{1}{d(A,C)}+\frac{1}{d(A,D)}\right)+d(A,B)\frac{2}{d(P,A)}\right)\end{aligned}\) Also, $d(P,A)+d(A,B)=d(P,B)$ and hence \(\begin{aligned} \Re(C,B;P,A)+\Re(D,B;P,A) &=\frac{d(P,A)}{d(P,B)}\left( d(A,B)\cdot \left(\frac{1}{d(A,C)}+\frac{1}{d(A,D)}\right)+2\left(\frac{d(P,B)}{d(P,A)}-1\right)\right)\\ &=\frac{d(P,A)}{d(P,B)}\left( d(A,B)\cdot \left(\frac{1}{d(A,C)}+\frac{1}{d(A,D)}\right)-2\right)+2\\ &=2S( d(A,B)/H-1)+2\end{aligned}\) where $H$ is the harmonic mean of $d(A,C)$ and $d(A,D)$, and $S=\frac{d(P,A)}{d(P,B)}$. Therefore, $d(A,B)=H$ if and only if $\Re(C,B;P,A)+\Re(D,B;P,A)=2$.

(ii) We use the previous result: $d(A,B)$ is the harmonic mean of $d(A,C)$ and $d(A,D)$ if and only if \(\Re(D,B;C,A)=\Re(C,B;C,A)+\Re(D,B;C,A)=2.\) (We make use of the convention that $\Re(C,B;C,A)=0$). Recall, from the discussion before Theorem [cocycleidentity]{reference-type=”ref” reference=”cocycleidentity”}, that \(\Re(D,B;C,A)=1-\Re(A,B;C,D).\) So \(\begin{aligned} \Re(C,B;C,A)+\Re(D,B;C,A)=\Re(C,B;C,A)+1-\Re(A,B;C,D)\end{aligned}\) and so the result follows by noting that the above quantity is equal to $2$ if and only if $\Re(A,B;C,D)=-1$. (This exercise was taken from @Eves:1972to [Theorem 2.6.4]).

4.12

10 centimetres. Let the image of the first foot of the yardstick be $AB$, the second foot be $BC$, the third foot be $CD$. Then \(\begin{aligned} \Re(A,B;C,D)&=\tfrac43\\ &=\mathrm{ratio}(D,B,A)/\mathrm{ratio}(C,B,A)=\\ &= \frac{35}{14}\mathrm{ratio}(D,B,A),\end{aligned}\) so \(\mathrm{ratio}(D,B,A)=\frac{8}{15}=\frac{d(D,C)+14}{d(D,C)+35},\) giving $d(D,C)=10$. (This exercise was adapted from [@Duffin:1993vt p. 42]).

4.13

By the Chasles-Steiner Theorem, since $P$, $Q$, $R_1$, $R_2$, $R_3$, $R_4$ are points on a conic, \(\Re(PR_1,PR_2;PR_3,PR_4)=\Re(QR_1,QR_2;QR_3,QR_4),\) but $\Re(PR_1,PR_2;PR_3,PR_4)=\Re(PS,PT;PU,PV)$ and \(\Re(QR_1,QR_2;QR_3,QR_4)=\Re(QT,QS;QV,QU)=\Re(QS,QT;QU,QV),\) so $\Re(PS,PT;PU,PV)=\Re(QS,QT;QU,QV)$. Therefore, by the Chasles-Steiner Theorem, $P$, $Q$, $S$, $T$, $U$, $V$ are points on a conic.