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Provide the exact list of steps needed to draw, using ruler and compass, a line $M$ through a given point $A$ and parallel to a given line $L$ (given by two points $B$ and $C$ on it). Assume that $A$ is not on $L$.


I am completely new to this. So what are we allowed to start off with? Do we start off with the line $L$ and points $B$ and $C$ on this line. Then we have a point $A$ that is not on this line. Then we need to use these starting points so that we get this line $M$?

I am still pretty unsure how to do this...

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  • $\begingroup$ how is this Galois theory? $\endgroup$ – cactus314 Mar 21 '16 at 23:33
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Draw the line perpendicular to $BC$, we'll call it $D$. Then draw the circle about $A$ that intersects $D$ twice, at points $E$ and $F$. You can find the midpoint of these two points, $G$. Drawing the line through $G$ and $A$ will give you a line perpendicular to $D$.

If you have three lines, $X$, $Y$, $Z$, and you know that $X$ and $Y$ are perpendicular to $Z$, can you say anything about how $X$ and $Y$ relate?

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  • $\begingroup$ I don't get how "Drawing the line through G and A will give you a line perpendicular to D".... it seems definitely not perpendicular. $\endgroup$ – snowman Mar 21 '16 at 23:46
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    $\begingroup$ Draw the triangle $EFA$. It's an isosceles triangle. Drawing the line straight down from $A$ to $EF$ will produce a point $G$ on $EF$. The line $AG$ will be at a right angle since it was drawn straight down. Since the triangle is isosceles, the length of $GE$ will equal the length of $GF$. Thus, $G$ is the midpoint, and we have the construction above. $\endgroup$ – Nikolas Wojtalewicz Mar 21 '16 at 23:52
  • $\begingroup$ where is your A initially? Is it anywhere that is not on line L? $\endgroup$ – snowman Mar 22 '16 at 12:25
  • $\begingroup$ also if we are not allowed to measure angles and lengths, how can we even draw a line perpendicular to BC? how would we know if it is right angled to BC? $\endgroup$ – snowman Mar 22 '16 at 12:32
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    $\begingroup$ youtube.com/watch?v=7Kp4P0fje_k $\endgroup$ – Nikolas Wojtalewicz Mar 22 '16 at 18:33
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The simplest way I know to construct line $M$ is to find the point $D$ such that $ABCD$ is a parallelogram. This can be done by intersecting the circle centred at $A$ with radius $p = BC$ and the circle centred at $C$ with radius $q = AB$.

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