What is this geometric principle?

Question

It appears that it does not matter how you move $A$, $B$, $C$, $D$, so long as $A$ and $D$ remain larger than $B$ and $C$, then you obtain the intersection between the three lines as shown ($y = x$, $[(A,B),(C,D)]$, $[(A,C),(B,D)]$).

Answer 1

Your construction can be rephrased in a more geometrical way as follows: given a triangle $ABC$, we construct a triangle $DBE$ having angle $\angle B$ in common with $ABC$ (in your case $\angle B$ is a right angle, but that is not necessary) and $AD=BD-BA=BC-BE=EC$ (see diagram below). Lines through $A$ and $D$, parallel to $BC$, meet lines though $E$ and $C$, parallel to $AB$, at $G$ and $F$ respectively. We want to prove that point $H$, where $AC$ and $DE$ meet, lies on line $FG$.

From point $H$ draw lines $HI$ and $HJ$, parallel to $BC$ and $AB$. By the intercept theorem, if we prove that $AI=EJ$ then $H$ lies on line $FG$.

Observe now that triangles $AIH$ and $HJC$ are similar, as well as triangles $DIH$ and $HJE$. We have then: $$ AI:HI=HJ:JC \quad\hbox{and}\quad DI:HI=HJ:JE, $$ whence: $$ HI\cdot HJ=AI\cdot JC=DI\cdot JE. $$ From that we get: $$ AI\cdot (JE+EC)=(AI+AD)\cdot JE, \quad\hbox{that is:}\quad AI\cdot EC=AD\cdot JE. $$ From the last equality, recalling that $EC=AD$, we obtain then $AI=JE$, which is what we wanted to prove.

Answer 2

The proof is easy with standard vector algebra: The line from (B,D) to (A,C) is parametrized by point coordinates (x,y) = (B,D)+t(A-B,C-D), 0<=t<=1, and the line from (C,D) to (A,B) is parametrized by point coordinates (x,y) = (C,D)+t'(A-C,B-D), 0<= t'<=1. The intersection between the two lines is found by equating the x and y coordinates of the two lines, solving B+t(A-B)=C+t'(A-C) together with D+t(C-D)=D+T'(B-D), which is a simple 2x2 linear system of equations. The solution is t=(D-B)/(A-C+D-B), t'=(D-C)/(A-C+D-B). Insertion of t and t' back into the line equations gives (B,D)+t(A-B,C-D) = [(AD-CB)/(A-C+D-B), (AD-CB)/(A-C+D-B)]. So indeed the intersection stays always on the diagonal because the x and y coordinates are always the same.