Proposition 3.9(a): If a ray r emanating from an exterior point of triangle ABC intersects
side AB at a point between A and B, then r also intersects side AC or side BC.
proof. (a) Let r= array XD with X an exterior point of triangle ABC and D ∈ AB such that
A ∗ D ∗ B (see first figure on next page).
(b) Line XD intersects either AC or BC. Without loss of generality, say line XD intersects AC at point E.
(c) Note E =/ A.
(d.5) Either E = C or A ∗ E ∗ C.
(f) Note C and E are on the same side of line AB
(g) Observe X and one vertex of the triangle must be on opposite sides of the line
through the other two vertices. Without loss of generality, say C and X are on
opposite sides of line AB.
(h) Exactly one of D ∗ E ∗ X, E ∗ D ∗ X, or D ∗ X ∗ E hold.
(i) Suppose E ∗ X ∗ D holds.
(j) Thus, E and X are on the same side of line AB
(K) Thus, C and X are on the same side of line AB
(l) This is a contradiction.
(m) So D ∗ E ∗ X or E ∗ D ∗ X hold.
(n) Therefore, r = ray XD intersects either AC or BC.
My solutions
(a) by Hypothesis
(b) Pasch’s theorem guarantees that the line containing the ray r (that is, the line XD in the picture) intersects either AC or BC.
Assume, for example, that it intersects AC at point E. We want to show that E is in the ray XD, that is, that ∼ E ∗ X ∗ D. Since every point between D and E is interior to ∆ABC but X is not, then, as we were hoping for, ∼ E ∗ X ∗ D.
(c) there exist a point E not on the line A, D, B (Proposition 2.3)
(d.5) if E=C, then E lies on line AC
(f) by definition Line AB and C,E point not on line AB, C, E are on the same side of line AB if C=E or if CE does not intersect line AB
(g) If C is on the same side of line AB as X, then C is on the opposite side from B, which means that line AB intersects BC and does not intersect AC; similarly if X is on the same side of AB as C, then AB intersects XD and does not intersect AC (separation axiom).
(h)we know by betweenness axion 3 that exactly one of the relations Exactly one of D ∗ E ∗ X, E ∗ D ∗ X, or D ∗ X ∗ E hold.
(I) Suppose E ∗ X ∗ D holds RAA hypothesis
(j) by definition AB is a line and E, X points not on line AB. E, X are on the same side of line AB if E=X or if XE does not intersect line AB.
(k) by definition AB is a line and C, X points not on line AB. C, X are on the same side of line AB if C=X or if XC does not intersect line AB.
(l) This is a contradiction (no idea what to write)
(m) by B-3 given any 3 distinct collinear points D,E,X exactly one is between the other two. So D ∗ E ∗ X or E ∗ D ∗ X hold.
(n) I had no idea what to write
Best Answer
Comments on comments :
(c) : "Note $E≠A$" we have to say that if $E=A$, then line $XD$ will intersect line $AB$ in two points : $D$ (which by construction is between A and B and thus $D≠A$) and $E$, contrary to the fact that two distinct lines have only one point in common.
(d.5) Due to the fact that $XD$ intersects $AC$ at point $E$, we have three possibilities: .Either $E = A$ or $E = C$ or $A ∗ E ∗ C$. The first one is excluded by (c); thus we are left with : $E = C$ or $A ∗ E ∗ C$.
(f) if C, E on different sides of line $AB$, then $CE=CA$ intersect $AB$ in a point Y such that $E ∗ Y ∗ C$; but $A$ is a point common to $AC$ and $AB$. Thus, or $CE=AB$ - impossible - or $Y=A$ - contrary to $E = C$ or $A ∗ E ∗ C$.
(i) Suppose E ∗ X ∗ D holds - i.e. we assume for contradiction the negation of what we are aimed to prove.
(l) from (k) we have that C and X are on the same side of line $AB$, but in (g) we have supposed that C and X are on opposite sides of line $AB$ : contradiction.
(m) thus, by RAA we have that $\lnot (E ∗ X ∗ D)$.
(n) we have shown that E is in the ray $XD$, by : $\lnot (E ∗ X ∗ D)$. By Pasch' theorem (b), $r =$ ray $XD$ intersects either $AC$ or $BC$.