Find the intersection of two lines entirely outside the given sheet of paper by straightedge alone This is a problem from Courant:

Two straight lines entirely outside the given sheet of paper are each
given by two pairs of straight lines intersecting at points of the
lines outside the paper. Determine their point of intersection by a
pair of lines through it.

Using Brianchon's theorem the solution (with all the lines/points on the paper) is: The two lines can be regarded as two of the three concurrent diagonals of the theorem. From the intersecting two pairs of lines a third concurrent diagonal can be constructed for each line. These two diagonals intersect in the point of intersection of the two original lines.
However, for this construction you need all intersection points, don't you?
Any idea? Or different approach?
 A: Define a “virtual point” $\tilde P$ to be a point, possibly off the paper, represented by a pair of lines $l, m$ on the paper passing through it. Define a “virtual line” $\tilde l$ to be a line, possibly off the paper, represented by a pair of virtual points $\tilde P, \tilde Q$ lying on it. We wish to prove that given two virtual lines $\tilde l, \tilde m$, we can construct their virtual point of intersection $\tilde A = \tilde l ∩ \tilde m$.
We’ll use Pappus’s hexagon theorem to prove three increasingly powerful lemmas, as follows.

Lemma 1. Given point $P$ and virtual point $\tilde Q$, we can construct the line $P\tilde Q$.
Proof. Say $\tilde Q$ is represented by $l, m$. Choose arbitrary points $B, E$. Draw $A = BP ∩ l$, $C = AE ∩ m$, $D = EP ∩ m$, $F = BD ∩ l$, $R = BC ∩ EF$, and $PR$. By Pappus’s theorem on hexagon $ABCDEF$, $P, \tilde Q, R$ are collinear, so $P\tilde Q = PR$.
If $P$ is on some line that intersects both $l, m$ on the paper, then by choosing $B, E$ sufficiently close to that line, we can perform this construction entirely on the paper. If not, we can instead perform it with an arbitrary $P'$ that is on such a line, yielding an auxiliary line concurrent with $l, m$, and repeat until we have auxiliary lines $l', m'$ suitable for the construction with $P$. □

Lemma 2. Given line $l$ and virtual line $\tilde m$, we can construct the virtual point $l ∩ \tilde m$.
Proof. Say $\tilde m$ is represented by $\tilde P, \tilde Q$. Choose arbitrary point $A$ and points $C, D$ on $l$. Using lemma 1, construct lines $A\tilde P, C\tilde Q, D\tilde P$, then let $E = AC ∩ D\tilde P$ and construct $E\tilde Q$. Draw $B = A\tilde P ∩ C\tilde Q$, $F = BD ∩ E\tilde Q$, and $AF$. By Pappus’s theorem on hexagon $ABCDEF$, the points $\tilde P, \tilde Q, l ∩ AF$ are collinear, so $l ∩ \tilde m$ is represented by $l, AF$.
We can always choose $A$ is sufficiently close to $l$ and $C, D$ on $l$ sufficiently close to $A$ that we can perform this construction entirely on the paper. □

Lemma 3. Given point $P$ and two virtual lines $\tilde l, \tilde m$, we can construct the line through $P$ concurrent with $\tilde l, \tilde m$.
Proof. Choose arbitrary points $B, E$. Using lemma 2, construct virtual points $\tilde A = BP ∩ \tilde l$, $\tilde C = AE ∩ \tilde m$, $\tilde D = EP ∩ \tilde m$, $\tilde F = BD ∩ \tilde l$. Using lemma 1, construct lines $B\tilde C$ and $E\tilde F$. Draw $R = B\tilde C ∩ E\tilde F$, and $PR$. By Pappus’s theorem on hexagon $\tilde AB\tilde C\tilde DE\tilde F$, $P, \tilde l ∩ \tilde m, R$ are collinear, so $PR$ is concurrent with $\tilde l, \tilde m$.
We can always choose $B, E$ sufficiently close to $P$ that we can perform this construction entirely on the paper. □

Theorem. Given two virtual lines $\tilde l, \tilde m$, we can construct the virtual point $\tilde l ∩ \tilde m$.
Proof. Choose two arbitrary points $P, Q$. Using lemma 3, construct lines $n$ through $P$ and $o$ through $Q$ concurrent with $\tilde l, \tilde m$. Then $\tilde l ∩ \tilde m$ is represented by $n, o$. □
A: Here I'd like to provide two brute-force approaches to solve most problems in this page:

Both approaches stretch / shrink invisible lines inside the sheet.
Homothetic / Similarity Transformation
Take the original question #24 as example, line AB and CD (red) are outside the sheet, where A, B, C and D are determined by QA-RA, QB-RB, QC-RC and QD-RD, respectively. Although A, B, C and D are outside the paper, we can construct PQ and PR by Q and R inside the sheet.

The steps are:

*

*Pick R' on QR close to Q, so that AB and CD can be shrunk inside the sheet;

*Construct parallels $R'A'{\parallel}RA$, $R'B'{\parallel}RB$, $R'C'{\parallel}RC$ and $R'D'{\parallel}RD$ (blue, green, yellow and purple);

*Construct $A'=QA{\cap}R'A'$, $B'=QB{\cap}R'B'$, $C'=QC{\cap}R'C'$ and $D'=QD{\cap}R'D'$;

*Construct A'B' and C'D', then $P'=A'B'{\cap}C'D'$;

*Construct the parallel $p{\parallel}PR$ (brown), then we can prove it easily that p passes through P. So PR is one of the line pair to determine P (brown);

*Repeat step 1-5 but change R' to Q' close to R, then we can get another line PQ (not shown in figure).


Perspective / Projective Transformation
As Anders Kaseorg mentioned, these problems are about projective geometry, where only straightedge is allowed during construction. So we should use perspective / projective transformation instead of homothetic / similarity transformation.
Firstly let's take question #20 as example:

The steps are:

*

*Pick AB on $l_1$ and CD on $l_2$, then take Q as the perspective center;

*Pick ${\triangle}A'B'C'$ close to Q, so that $H=AC{\cap}A'C'$ and $J=BC{\cap}B'C'$ on their perspective axis (applying Desargues's theorem on ${\triangle}ABC$ and ${\triangle}A'B'C'$) are inside the sheet;

*Construct $K=AD{\cap}HJ$, then $D'=A'K{\cap}QD$;

*Construct $P'=A'B'{\cap}C'D'$, then P'Q passes through P, because points ABCDP and points A'B'C'D'P' are perspective about center Q.

This approach is more complicated than Anders Kaseorg's proof to Lemma 1, but it can apply to any other cases, such as #24:

The steps are:

*

*Pick E and E' on QA, F and F' on QB, R' on QR, that are inside the sheet;

*Take Q as the perspective center, then construct the perspective axis GH by applying Desargues's theorem on ${\triangle}EFR$ and ${\triangle}E'F'R'$;

*Construct lines JR'A', KR'B', LR'C' and MR'D' (dash lines) which are perspective to lines JRA, JRB, JRC and JRD (solid lines) about Q (note that QA, QB, QC and QD (dot lines) are unchanged because Q is the perspective center);

*Construct $P'=A'B'{\cap}C'D'$, then P'Q passes through P (brown dot line), because points ABCDP and points A'B'C'D'P' are perspective about center Q;

*Repeat step 1-4 but take R as perspective center, then we can get another line PQ (not shown in figure).


Assumption
For $A=l_A{\cap}m_A$, $B=l_B{\cap}m_B$, $C=l_C{\cap}m_C$ and $D=l_D{\cap}m_D$ in problem #24, we assume that $l_A$, $l_B$, $l_C$ and $l_D$ meet at Q, and $m_A$, $m_B$, $m_C$ and $m_D$ meet at R, and both Q and R are inside the sheet.
What should we do if these 4 pairs of lines are not meet at Q and R?
Just pick two arbitrary points Q and R, then construct QA, QB, QC, QD, RA, RB, RC and QD, by applying the result of problem #20 or Anders Kaseorg's Lemma 1.
