2.3

We will use the fact that the collineation group of $\mathsf{PG}(2,\mathbb{R})$ acts transitively on pairs $(P,\ell)$ where $P$ and $\ell$ are a point and line that are not incident. This allows us to suppose without loss of generality that $C=(0,0,1)$ and $\ell:z=0$. It is a routine calculation that the subgroup of $\mathsf{PGL}(3,\mathbb{R})$ fixing $C$ line-wise and $\ell$ point-wise consist of collineations induced by the matrices of the form \(\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&\gamma \end{bmatrix},\) where $\gamma\in\mathbb{R}\backslash\{0\}$. Clearly this group is isomorphic to $\mathbb{R}\backslash\{0\}$ under multiplication, which is abelian. So two homologies with centre $C$ and axis $\ell$ lie in this group and thus commute.

(We remark that this proof can also be done synthetically, but one needs to use the fact that Pappus’ Theorem holds in $\mathsf{PG}(2,\mathbb{R})$).

2.4

(a) (We assume that the product of two central collineations is a central collineation). First, notice that $\phi\psi$ fixes the point $C$ as $C^{\phi\psi}=(C^\phi)^\psi=C^\psi=C$. Now let $\ell$ be a line incident with $C$. Then $\phi$ fixes $\ell$ and $\psi$ fixes $\ell$. Therefore, $\phi\psi$ fixes $\ell$. So $\phi\psi$ is an elation with centre $C$.

(b) Let $u$ be the line on the two points $a\cap m^\tau$ and $b\cap m^\sigma$. Now $m^\tau$ is the line joining $b\cap m$ and $a\cap u$. So \(\begin{align*} m^{\tau\sigma}&=(b\cap m)^\sigma(a\cap u)^\sigma\\ &=(b\cap m^\sigma)(a\cap u^\sigma)\\ &=u,\end{align*}\) as $a\cap u^\sigma=a\cap u$. Similarly, $m^{\sigma\tau}=u$ and hence $m^{\sigma\tau}=m^{\tau\sigma}$.

(c) Let $\psi$ and $\phi$ be two elations sharing a common centre $C$. From (a), $\psi\phi$ and $\phi\psi$ are both elations with centre $C$. Let $\ell$ be the axis of $\psi\phi$ and let $X$ be a point incident with $\ell$. Let $m$ be a line. If $m$ is incident with $C$, then $m^{\psi\phi}=m=m^{\phi\psi}$. Otherwise, if $m$ is not incident with $C$, then by (b), we also have $m^{\psi\phi}=m=m^{\phi\psi}$. Therefore, $\psi\phi \psi^{-1}\phi^{-1}$ fixes every line of the projective plane and so is equal to the identity.

2.11

Apply Pappus’ Theorem to $A$, $m_1\cap n_2$, $m_1\cap n_3$, $A’$, $n_1 \cap m_3$, $n_1 \cap m_2$. Then $A(n_1 \cap m_2)=m_2$, $A’(m_1 \cap n_3)=n_3$, $A(n_1 \cap m_3)=n_1$, $A’(m_1\cap n_2)=n_2$. So $(m_1\cap n_2)(m_2 \cap n_1)$, $(m_3 \cap n_1)(m_1 \cap n_3)$ and $(m_3 \cap n_2)(m_2 \cap n_3)$ are concurrent.

2.12

Apply Pappus’ Theorem to $AB \cap CD$, $F$, $G$, $AD \cap BC$, $H$, $I$. Then $(AB \cap CD)H=AB$, $(AD \cap BC)F=BC$ and these lines meet at $B$. So $(AB \cap CD)I=CD$ and $(AD \cap BC)G=AD$, and these lines meet at $D$. Hence $B$, $D$ and $FI \cap GH$ are collinear.

2.14

The triangles $AFH$, $UCG$ have $AF \cap UC=D$, $AH \cap UG=V$, $FH \cap CG=B$, with $D$, $V$, $B$ collinear, so, by the dual of Desargues’ Theorem, $AU$, $FC$ and $HG$ are concurrent.

2.15

The triangles $XYZ$, $DCB$ are in perspective from $A$ (and $XYZ$ is a triangle by Fano’s axiom), so, by Desargues’ Theorem, the intersections $L$, $M$, $N$ of corresponding sides are collinear.

2.16

Apply Desargues’ Theorem to the triangles $AED$ and $BFC$, which are in perspective from $P$. This shows that the intersection points of the diagonals of quadrilaterals $ABCD$, $ABEF$ and $CDFE$ lie on a line. Now apply Desargues’ Theorem to the triangles $AFC$ and $BED$, which are in perspective from $P$. This shows that the diagonals of quadrilaterals $ABCD$ and $CDFE$ and $Q$ are collinear.

2.17

We have $B’D\parallel AF \parallel A’E$ and $BD \parallel B’F \parallel EC’$ and $AE \parallel FA’ \parallel DC’$. Now work in $\mathsf{PG}(2,\mathbb{R})$. Let $A’’$ be the point at infinity on $FA’$, $B’’$ be the point at infinity on $FB’$ and $C’’$ be the point at infinity on $DB’$. Then $A”F \cap C”E=A’$, $B”F \cap C”D=B’$, $A”D \cap B”E=C’$. So, by Pappus’ Theorem, $A’$, $B’$, $C’$ are collinear.

2.18

The six lines $a_1:=BC$, $a_2:=AG$, $a_3:=EF$ and $b_1:=AB$, $b_2:=GF$, $b_3:=DC$ form two parallel triples, so the dual of Pappus’ Theorem applies: $\langle a_1 \cap b_2, a_2 \cap b_1\rangle=HA$, $\langle a_1 \cap b_3,a_3 \cap b_1\rangle=CE$ and $\langle a_2 \cap b_3,a_3 \cap b_2\rangle=DF$ are concurrent.

2.19

Let the diagonals of quadrilaterals $ABCD$, $ABEF$ and $CDFE$ intersect respectively at $R,S$ and $T$. Triangles $ACF$ and $BDE$ are in perspective from $P$, so, by Desargues’ Theorem, $R$, $S$ and $T$ are collinear. Triangles $ADE$ and $BCF$ are in perspective from $P$, so, by Desargues’ Theorem, $Q$, $S$ and $T$ are collinear.