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.

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    $\begingroup$ Not a perfect answer - perhaps someone else could develop it into something better. Drop perpendiculars from $B$ to the line segments (hoping that they lie on the page), and connect the intersection points, say $C$ and $D$. Now $ABCD$ is a cyclic quadrilateral, so angle $ADC$ = angle $ABC$. So, copy angle $ADC$ to $B$, and hence we can construct $E$ such that angle $EBC$ = $ABC$, so $A$, $E$, $B$ collinear. $\endgroup$ – B. Mehta Nov 28 '17 at 18:45
  • $\begingroup$ This has issues in that copying an angle might take many steps (I know it's possible, I don't recall how) and that $C$ and $D$ may not lie on the page, let alone be part of the original line segments. $\endgroup$ – B. Mehta Nov 28 '17 at 18:46

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.

enter image description here


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.

enter image description here

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 $(H\equiv B)$.

  • $\begingroup$ This is a great answer. But there is a problem, it doesn't work for any point B on the page without allowing you to extend $L_1$ and $L_2$ to find an intersection with $L_1$ and $L_2$ with their parallels. However, I suppose I didn't say this wasn't allowed explicitly in my question, so I can't fault you for this. $\endgroup$ – William Oliver Nov 28 '17 at 19:57
  • $\begingroup$ For example, it is possible to construct a situation where $D$ is actually past $A$ by moving $B$ straight up, outside of the two original line segments, until $BD$ intersects $L_2$ at a point beyond $A$. $\endgroup$ – William Oliver Nov 28 '17 at 20:14
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    $\begingroup$ Of course, you are right. But this was just to give a principle, based on the property of parallelogram. In the general case, one can construct an homothetic figure at any smaller scale that wanted. So that, all will remain into the limits of the page. This gives the direction (slope) of the line AB that we are looking for. $\endgroup$ – JJacquelin Nov 28 '17 at 21:51

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.

enter image description here


Consider the following figure:

enter image description here

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 $A$.


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).

enter image description here


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