Applied projective geometry

Trying to draw a horizontal like without a level

One day I was too lazy to go get a level and didn’t want to measure. So, I tried to apply projective geometry to draw one with a no-compass, straightedge-only construction.

Two different parallel lines and a point not on them are given. Using only straightedge construct a line passing through the given point that is parallel to the two given lines.

Applied projective geometry

Solution

Let the two given lines be called $\alpha$ and $\beta$, and the given point $P_1$. First we draw two points $A_1$ and $A_2$ on $\alpha$ and learn how to construct the midpoint of the segment $A_1A_2$.

Constructing midpoint of a segment:

  1. Take a point $B_1$ on $\beta$ and draw the line $A_1B_1$.
  2. Take a point $C$ on $A_1B_1$, that is outside $\alpha$ and $\beta$.
  3. Draw the line $CA_2$ and let $B_2$ be the point of intersection of this like with $\beta$.
  4. Draw the lines $A_1B_2$ and $A_2B_1$. Let $D$ be the point of intersection of these lines.
  5. Draw the line $CD$. The point of intersection of this line and the line $\alpha$ is the midpoint of $A_1A_2$.

Construction of the third parallel line:

  1. Select to points $A_1$ and $A_2$ on $\alpha$.
  2. Construct the midpoint $A_3$, of the segment $A_1A_2$. [The line $\beta$ is no longer needed.]
  3. Draw the line $A_1P_1$.
  4. Select a point $Q$ on the line $A_1P_1$, different from $A_1$ and from $P_1$.
  5. Draw the lines $QA_2$ and $QA_3$.
  6. Draw the line $A_2P_1$ and let $R$ be its intersection with $QA_3$.
  7. Draw the lines $A_1R$ and $QA_3$ and their intersection $P_2$.
  8. The line $P_1P_2$ is parallel to the line $\alpha$.