Proof Portfolio
Table of Contents
- Chapter 7
- Proof Theorem 7.2
- Proof Theorem 7.7
- Proof Exercise 7.3.18
- Proof Lemma 7.1
- Proof of Independence of the Dual Axioms
- Euclid's Elements
- Betweenness Proposition I
- Isosceles triangle proof of congruent base angles
- Chapter 3
- Proof of Hypotenuse-Leg Criterion for congruence of right triangles
- 3.3.42 (Perpendicular bisectors concurrence proof)
Proof Theorem 7.2
7.2: Each point in an affine geometry of order k lies on k+1 lines.
Proof: [br]Consider any line [i]l[/i] in an affine plane. By Theorem 7.1, we know that there are k distinct points that lie on [i]l.[/i] Now, consider the point P not on [i]l. [/i]Axiom 2 tells us that any two distinct points have exactly one line in common. There are k points on line l distinct from P. Thus, there are at least k distinct lines through point P. By Axiom 3, we also know that there is a line through point P that does not intersect line [i]l[/i]. Thus, there are at least k+1 distinct lines through point P. Now, consider that there are k+2 lines through point P. By Axiom 1, this would imply that there is a point O distinct from those that lie on line [i]l, [/i]and that does not lie on the line through P parallel to [i]l. [/i]We then know that the line through point P and O must also intersect line [i]l [/i]at some point R because Axiom 3 tells us that there is exactly one line through point P not intersecting [i]l. [/i]Either this point R is distinct from the other k distinct points on [i]l[/i] or point R is one of those k distinct points. If R is one of the k distinct points then we still have k+1 distinct lines through point P. If R is not one of the k distinct points then there are now k+1 points on line [i]l [/i]and we no longer have order k. Thus, there must be exactly k+1 lines through point P. Without loss of generality, we can then say that given any point in an affine geometry of order k there are exactly k+1 lines through that point. Or, each point in an affine geometry of order k lines on k+1 lines.
Betweenness Proposition I
Proof: Consider the four points A,B,C,D. Assume A*B*C and B*C*D. By Proposition of line separation, A*B*C implies A and C are on opposite sides of the point B. Similarly, B*C*D implies that C and D are on the same side of B. Thus, because C is on the opposite side of B from A and C is on the same side of B as D, we know that A is on the opposite side of B as D. That is, A*B*D. Again, by Proposition of line separation, A*B*C implies that A and B are on the same side of C. Similarly, B*C*D implies that B and D are on opposite sides of C. Thus, as before, we can infer that A and D are on opposite sides of C. That is, A*C*D. Now assume A*B*D and B*C*D. By the Proposition of line separation, A*B*D implies that A and D are on opposite sides of B and B*C*D implies that C and D are on the same side of B. Thus, A and C must be on opposite sides of B. That is, A*B*C. Similarly, A*B*D implies that A and B are on the same side of D and B*C*D implies that B and C are on the same side of D. Thus, A and C are on the same side of D. That is, A*C*D. [br]
Proof of Hypotenuse-Leg Criterion for congruence of right triangles
"Given two triangles, [math]\triangle ABC[/math] and [math]\triangle DEF[/math] with right angles at [math]\angle A[/math] and [math]\angle D[/math], [math]\overline{AB}\cong\overline{DE}[/math], and [math]\overline{BC}\cong\overline{EF}[/math]. Prove that [math]\triangle ABC\cong\triangle DEF[/math]."[br][br][b]Proof:[/b][br]Consider the point X, on the line [math]\overline{AC}[/math] such that X is on the opposite side of A from C and such that [math]\overline{AX}\cong\overline{DF}[/math]. By Euclid's Proposition 13, [math]\angle BAX[/math] is a right angle. Thus, [math]\angle BAX\cong\angle EDF[/math]. We also know that [math]\overline{AB}\cong\overline{DE}[/math]. Therefore, we have satisfied the SAS criterion for congruent triangles. That is, [math]\triangle ABX\cong\triangle DEF[/math]. [br][br]Notice then, that [math]\overline{BX}\cong\overline{EF}\cong\overline{BC}[/math]. Thus, [math]\triangle BCX[/math] is an isosceles triangle. Therefore, [math]\angle C\cong\angle X\cong\angle F.[/math] Then, the AAS criterion for congruent triangles is satisfied for [math]\triangle ABC[/math] and [math]\triangle DEF[/math]. So, we can conclude that [math]\triangle ABC\cong\triangle DEF[/math].