Methods for Mascheroni Constructions

To construct a circle at $A$ having a given radius $BC$ with a collapsing compass.

$A_{AB},B_{BA}$	$D_{DC},E_{EC}$	$A_{AF}$
$D$, $E$	$C$, $F$

To construct the reflection, $Y$, of $X$ with respect to line $AB$.

$P_{PX},Q_{QX}$	$Y$
$X$, $Y$

To construct a segment $n$ times as long as given segment $AB$, where $n$ is a positive integer

$A_{AB},B_{BA}$	$C_{CA},B_{BA}$	$D_{DC},B_{BA}$	$N$
$C$ and some other point	$A$, $D$	$C$, $N$

$AN = 2AB$. Continuing in this manner one can construct $3AB$, $4AB, \dots$.

To construct ST, the fourth proportional to $a,b,c$.

$O_a,L_c$	$L_t,O_b$	$M_t,O_b$	ST
($L$ on $a$) M	$S$ in the interior of LOM	$T$ not in the interior of $\angle LOM$

If $c>2a$ then by taking $n$ large enough one can make $2na>c$. Now use the above method with $na$, $nb$, and $c$.

To construct the midpoint F of an arc AB with center O.

$A_{AO},O_{AB}$	$B_{BO},O_{AB}$	$C_{CB},D_{DA}$	$C_{OE}$, arc $AB$
$C$	$D$	$E$	$F$

Operation $A$, to draw a circle with a given center and radius, and Operation $B$, to find the points of intersection of two circles, are two operations that can clearly be accomplished with only a compass. To accomplish Operation $C$, to find the points of intersection of a line and a circle, use (1) to reflect circle $O_r$ through line $AB$ to circle $O^\prime_r$. The intersection points of $O_r$ and $O^\prime_r$ are the intersection points of line $AB$ and $O_r$. If the reflection of $O$ is $O^\prime$ ($O$ lies on line $AB$), then choose any point $P$ on the circle and use (1) to reflect it through line $AB$ to $Q$ which will also lie on circle $O$. Use (4) to find the midpoints of major and minor arcs $PQ$. These midpoints are the points of intersection of line $AB$ and $O_r$. If the reflection of $P$ is $Q$ then $P$ is one of the points on $AB$ and $O_r$. To find the other point use (2) to step around circle $0$.
To accomplish Operation $D$, to find the point of intersection, $F$, of two lines $AB$ and $CD$, first use (1) to reflect $C$ and $D$ through line $AB$. Then complete parallelogram $CC^\prime ED$ by drawing $C^\prime_{CD}$ and $D_{CC^\prime }$. Note that $DD^\prime E$ is a straight line. Since $\Delta DD^\prime F \sim \Delta E D^\prime C^\prime$, use (3) to construct the fourth proportional $x$ to $D^\prime E, DD^\prime$, and $C^\prime D^\prime$. The intersection of $D_x$ and $D^\prime_x$ is $F$.