How to intersect a line with a point off the page using a straight edge and compass (see description)

Question

Its very common in illustration to want to draw a line towards a vanishing point that is off of the page.

The specific problem is this: lets say we draw two line segments on a piece of paper lying on a flat surface. The line segments are drawn such that, if extended beyond the edge of the page, they intersect at a point $A$. Next, we draw a point $B$ anywhere on the page.

Without actually extending the two line segments to find $A$, is there a way to draw a line segment such that, if extended to a line, would intersect both $A$ and $B$ using only a straight edge and compass construction?

Looking for an answer with the fewest number of steps.

Answer 1

On the graph below :

The two given lines are $(L_1)$ and $(L_2)$.

From B, draw a parallel to $(L_1)$. It intersect $(L_2)$ in D.

From B, draw a parallel to $(L_2)$. It intersect $(L_1)$ in C.

With straight edge and compass, draw the centre M of the segment CD.

Draw the line BM.

The point A is on this line, but outside the page.

$$ $$ METHOD WITH HOMOTHETY :

Chose an arbitrary homothetic ratio $1/n$ with $n$ large enough so that the figure becomes sufficiently small to be entirely on the page.

For example, on the next drawing, $n=3$.

$(L_1)$ and $(L_2)$ are the two given straight lines which the intersection point A is outside the page.

With center of homothety $H$, draw the homothetic line of $(L_1)$, that is :

Take an arbitrary point P on $(L_1)$ and divide HP in $n$ equal parts.

The method to divide a segment in equal parts is well known. Just to remind it, the accessory construction in yellow on the figure : Draw $n$ aligned equal segments of arbitrary length
Hh$_1=$h$_1$h$_2$=...=h$_{n-1}$h$_n$. From h$_1$ , h$_2$ , ... , h$_{n-1}$ draw the parallels to h$_n$P. They intersect HP in p$_1$ , p$_2$ , ... , p$_{n-1}$ . So that we have Hp$_1=$p$_1$p$_2$=...=p$_{n-1}$P.

Take an arbitrary point Q on $(L_)$ and divide HQ in $n$ equal parts with the same method. Hq$_1=$q$_1$q$_2$=...=q$_{n-1}$Q.

Then, from p$_1$ draw the parallel to (L$_1$). From q$_1$ draw the parallel to (L$_2$). They intersect in A'.

Draw the straight line HA'. The point A is on this line, outside the page.

Note : The point q$_1$ is not on (L$_1$). It is just accidentally that it appears close to (L$_1$).

Note : The center of homothety $H$ was taken at point $B$ as given in the question $(H\equiv B)$.

Answer 2

Consider the following figure:

The lines, $a$ and $a'$ are given and intersect each other at $A$ which is not on the paper.

Construct $b$ and $b'$ so that their distance from $a$ and $a'$ are the same and their intersection point (white $A$) be on the paper. We get a white parallelogram centered at $X$. From the white $A$ through $X$ we get a line that goes through $A$.

Now, draw through $B$ a parallel to $XA$. From $B$ on the latter line construct (backwards) a segment whose length is the double of the segment white $AX$.

Let the end point of that segment be denoted by $B'$. Draw a line through $B'$ and white $A$. Finally draw a parallel to this line through $B$. This line will go through $A$.

Answer 3

The following solution to the problem uses the ruler only and is based on Desargues's theorem.

Let $R=AB\cap A'B'$ be the point out from the sheet and $P$ a point on the sheet. It's required to draw the line $PR$.

Let $O=AA'\cap BB'$, and choose a line $c\supset O$. Let $C=c\cap AP$ and $C'=c\cap A'P$. The triangles $ABC$ and $A'B'C'$ are homologous since $O=AA'\cap BB'\cap CC'$, thus $R=AB\cap A'B'$, $P=AC\cap A'C'$ and $Q=BC\cap B'C'$ lies on a line. Consequently, $PQ$ is the required line.

Answer 4

An idea with reflection of the off-the-page point $A$ with respect to some "suitable" line $BG$, assuming given lines $CD$ and $EF$, the sought line segment is $BN'$ (4 circles and 5 lines).