Proofs Portfolio
Table of Contents
- Chapter 2
- Proof: Betweenness: Proposition 1
- Proof: Isosceles Triangle - Congruent Base Angles
- Chapter 3
- Proof: AAS Criterion
- Proof: Hypotenuse-Leg Criterion
- Proof: Perpendicular Bisector Concurrence
- Prove: Exercise 3.22
- Proof: Converse of Pythagorean Theorem
- Proof: Ceva's Theorem
- Chapter 4
- Proof: Theorem 4.5
- Proof: Theorem 4.1
- Chapter 5
- Proof: 5.2
- Proof: 5.31
- Proof: 5.25
- Chapter 6
- Proof: 6.3.4
- Chapter 7
- Proof: Theorem 7.2
- Proof: Theorem 7.7
- Proof: Lemma 7.1
- Proof: Dual Axioms Independence
- Chapter 8
- Proof: 8.34
- Proof: 8.3,4,5
- Chapter 11
- Proof: 11.5.12
- Applet - Final Proof
- Nine-point Circle
Proof: Betweenness: Proposition 1
Between: Proposition 1
Using the Axioms of Independence and betweenness and the line separation property, show that sets of four points A, B, C, D on a line behave as we expect them to with respect to betweenness. Namely, show that:[br]1) A*B*C and B*C*D imply A*B*D and A*C*D[br]2) A*B*D and B*C*D imply A*B*C and A*C*D[br][br]1) Consider a line, [math]l_1[/math], with points A, B, C, D on the line in the following order: A*B*C and B*C*D. By Axiom B1, we know that B lies between points A and C, while C lies between B and D. Note that B separates the line segment AD into two sets: one set containing A, and one set containing C and D. Therefore, because of the order of our points, we can conclude that A*B*D. Similarly, we can use C to separate segment AD where one set contains points A and B, and the other set contains D. Thus, we have A*C*D.[br][br]2) Consider a line, [math]l_1[/math], with points A, B, C, D on the line in the following order: A*B*D and B*C*D. By Axiom B2, we know that B lies between points A and D, while C lies between B and D. Note that B separates the line segment AD into two sets, one set contain A, and one set containing C and D. Similarly to part 1, we can conclude that A*B*C. Because we can also separate segment AD into two sets using point C, we can then use similar logic to determine that A*C*D.[br]
Proof: AAS Criterion
Proof: AAS Criterion
The angle-angle-side criterion for congruent triangles says that: If two angles and a non-included side of one triangle are congruent respectively to two angles and a non-included side of another triangle, then the two triangles are congruent.[br][br]Given triangles [math]\Delta[/math]ABC and [math]\Delta[/math]DEF with [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D, [math]\angle[/math]C [math]\cong[/math] [math]\angle[/math]F, and BC [math]\cong[/math] EF. Prove that[br][math]\Delta[/math]ABC and [math]\Delta[/math]DEF.[br][br]Proof:[br]Let [math]\Delta[/math]ABC and [math]\Delta[/math]DEF with [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D, [math]\angle[/math]C [math]\cong[/math] [math]\angle[/math]F, and BC [math]\cong[/math] EF. Prove that [math]\Delta[/math]ABC [math]\cong[/math] [math]\Delta[/math]DEF. [br][br]Note: If AC [math]\cong[/math] DF, then the triangles would be congruent because of SAS:[br](AC [math]\cong[/math] DF, BC [math]\cong[/math] EF, [math]\angle[/math]C [math]\cong[/math] [math]\angle[/math]F)[br][br]Case 1: Assume AC < DF. Then by (III-1), there is a point X on DF such that AC [math]\cong[/math] XF. Then[br]ABC [math]\cong[/math] XEF. So [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]X, but this isn't the case because we assumed [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D. [br]Thus, AC is not less than DF. [br][br]Case 2: Assume AC > DF. They by (III-1), there is a point X on AC such that XC [math]\cong[/math] DF. [br]Then XBC [math]\cong[/math] DEF. So [math]\angle[/math]X [math]\cong[/math] [math]\angle[/math]D, but this isn't the case because we assumed [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D. Thus, AC is not greater than DF. [br][br]Since AC is neither less than or greater than DF, it must be equal to DF. Therefore, ABC [math]\cong[/math] DEF. [br]
Proof: Theorem 4.5
Proof: Theorem 4.5
Theorem 4.5: If three circles are given with noncollinear centers and each pair of circles determines a radical axis, then three radical axes are concurrent.[br][br]Proof: Assume A, B, and C are circles with non-collinear centers. Let [math]x_1[/math] be the center of A, [math]y_1[/math] be the center of B, and [math]z_1[/math] be the center of C. Let B intersect both A and C. In order to find the radical axes of each pair of circles, we must first determine the points of intersection between all circles. Note that the points of intersection for circles A and B are points m and n. Similarly, the points of intersection between B and C are points o and p. By definition of radical axis, we know that line [math]l_1[/math] is the radical axis of circles A and B. Similarly, we know that [math]l_2[/math] is the radical axis of circles B and C. Construct point X which is the intersection of [math]l_1[/math] and [math]l_2[/math]. Since circles A and C do not intersect, start by constructing line t through [math]x_1[/math] and [math]z_1[/math]. Then construct line [math]l_3[/math] that is perpendicular to t and passes through point X. Thus, all radical axes pass through point X. Thus, [math]l_1[/math], [math]l_2[/math], and [math]l_3[/math] are concurrent.
Proof: 5.2
Proof: 5.2
Using coordinates, write a detailed step-by-step proof that the set of points equidistant from two fixed points, A and B, is the perpendicular bisector of the segment AB.[br][br]Let AB be a line segment with endpoints A and B. Let C be the midpoint of AB such that C is centered at the origin. Thus, we know that C = (0, 0). We will denote A and B as follows: A = (x, 0) and B = (-x, 0).[br][br]Now, let D be another point which is equidistant from line segment AB. Denote D as [math](x_1,y_1)[/math].[br]Using the distance formula, we see that:[br][br]1) [math]dist\left(A,D\right)=\sqrt[]{\left(x-x_1\right)^2+\left(0-y_1\right)^2}[/math][br]2)[math]dist\left(B,D\right)=\sqrt[]{\left(-x-x_1\right)^2+\left(0-y_1\right)^2}[/math][br][br][br]Since we know that D is equidistance from both A and B, we can say that dist(A,D) = dist(B,D).[br][br]Consider the following:[br][br][math]\sqrt[]{\left(x-x_1\right)^2+\left(0-y_1\right)^2}=\sqrt[]{\left(-x-x_1\right)^2+\left(0-y_1\right)^2}[/math][br][math]\Longrightarrow\left(x-x_1\right)^2+\left(0-y_1\right)^2=\left(-x-x_1\right)^2+\left(0-y_1\right)^2[/math][br][math]\Longrightarrow\left(x-x_1\right)^2=\left(-x-x_1\right)^2[/math][br][math]\Longrightarrow\left(x-x_1\right)^{ }=\left(-x-x_1\right)^{ }[/math][br][math]\Longrightarrow2x=x_1-x_1[/math][br][math]\Longrightarrow x=0[/math].[br][br]Thus, for any point D that we choose equidistant from A and B, the x coordinate must always be zero. We know that C is the midpoint of AB, and the x coordinate of C is zero, thus lying on the y-axis. Note that by definition, the y-axis is perpendicular to the x-axis. Thus, any point D equidistant from A and B forms the line CD which is perpendicular to AB since point points lie on the y-axis. Therefore, the set of all points equidistant from AB must lie on line CD, the perpendicular bisector of AB.[br][br][br][br][br]
Proof: 6.3.4
Proof: 6.3.4
Let g(P, Q) = max([math]x_P,x_Q[/math]) + max([math]y_P,y_Q[/math]).[br][br]A) Prove or disprove that g is a metric:[br][br][b][u]Counter Example:[/u][/b][br][br][b][u]Metric Axiom 1:[/u][/b] Let P and Q be points where P = (1, 2) and Q = (1, 2). Thus, P = Q.[br][br]Notice that: g(P, Q) = max(1, 1) + max(2, 2) = 1 + 2 = 3. Obviously, we can see this fails Metric Axiom 1 because g(P, Q) [math]\ne[/math] 0.[br][br]Therefore, g does not define a metric.[br][br][b][u][color=#ff0000]The rest of this is for communication only and not part of the proof: [/color][/u][/b][br][br][b][u]Metric Axiom 2:[/u][/b] Let P = (a, b) and Q = (x, y). We need to show that g(P, Q) = g(Q, P).[br][br]Notice that: g(P, Q) = max(a, x) +max(b, y) and g(Q, P) = max(x, a) + max(y, b). In both cases, 'max' will choose the greater value of each pair which will be the same number regardless of the individual coordinates. Therefore, g(P, Q) = g(Q, P).[br][br][br][b]Metric Axiom 3:[/b] Let P, Q, and Z be points where P = (a, b), Q = (c, d), and Z = (e, f).[br][br]Notice that g(P, Q) + g(Q, Z) = max(a, c) + max(b, d) + max(c, e) + max(d, f) [math]\ge[/math] g(P, Z) = max(a, e) + max(b, f). This is the case because the 'max' function always chooses the largest of two values. Since g(P, Q) + g(Q, Z) is the sum of four 'max' functions, we know that its sum must always be greater than the sum of two 'max' functions containing on the same set of values: {a, b, c, d, e, f}. [br][br][br]
Proof: Theorem 7.2
Theorem 7.2
Theorem 7.2: Each point in an affine plane of order [math]k[/math] lies on [i][math]k+1[/math][/i] lines. [br][br]Proof: Let [math]l_1[/math] be an arbitrary line. By Theorem 7.1, we know that [math]l_1[/math] has k points. Let [i]p[/i] be an arbitrary point not on [math]l_1[/math]. By Axiom 2, every point on [math]l_1[/math] must have exactly one line in common with [i]p[/i]. By Axiom 3, [i]p[/i] has exactly one line parallel to [math]l_1[/math]. [br][br]Therefore, any point in an affine plane of order [math]k[/math] has [math]k+1[/math] lines.
Proof: 8.34
Develop a transformational proof that the base angles of an isosceles triangle must be congruent.
[b][u]Proof:[/u][/b] Consider an isosceles triangle ABC. By definition, we know that [math]\overline{AB}\cong\overline{BC}[/math]. [br][br]First, construct the perpendicular bisector of AC, which we will call [math]l_1[/math]. Let Z be the point at which [math]l_1[/math] intersects AC. As consequence of triangle ABC being isosceles, we know that [math]l_1[/math] will pass through point [math]B[/math].[br][br]Then, reflect triangle ABC about [math]l_1[/math]. As result of a reflection being an isometry, we know that it preserves angle measurements and distances between points. When triangle ABC is reflected, notice that BC maps to AB because [math]\overline{AZ}\cong\overline{CZ}[/math] by definition. Consequently, note that angles [math]\angle A[/math] and [math]\angle C[/math] switch places. However, also note that [math]\angle B[/math] remains in its original position because all points along [math]l_1[/math] are fixed regardless of the reflection. Since the distances and angles in this triangle are preserved, we know that when ABC is reflected, the triangle formed is congruent to ABC. For clarity, note that since the two segments which define angle A map to the two segments which define angle C, we know that [math]\angle A\cong\angle C[/math].
Proof: 11.5.12
Prove that if the three perpendicular bisectors of the sides of [math]\Delta[/math]ABC in the hyperbolic plane are concurrent at a point O, then the circle with center O and radius OA also contains the points B and C. Prove that this circumcircle is unique.[br][br][b][u]Proof:[/u][/b][br][br]([math]\Longrightarrow[/math])[br][br]Let [math]\Delta[/math]ABC be a triangle in the hyperbolic plane such that the three perpendicular bisectors of the sides of [math]\Delta[/math]ABC are concurrent at point O. Let [math]l_1[/math] be the perpendicular bisector of AC which intersects AC at point X, [math]l_2[/math] be the perpendicular bisector of BC which intersects BC at point Y, and [math]l_3[/math] be the perpendicular bisector of BC which intersects BC at point Z. We wish to show that points A, B, and C lie on a common circle, E, centered at O with radius OA. Thus, we need to show that [br]OA = OB = OC.[br][br]First, construct line segments OA, OB, and OC. Note that X is the midpoint of AC because [math]l_1[/math] is the perpendicular bisector of AC. From this, we know that [math]\angle[/math]AXO = 90 = [math]\angle[/math]CXO. Furthermore, we know that OX = OX and AX = CX. Therefore, by SAS congruence, we know that [math]\Delta[/math]AXO [math]\cong[/math] [math]\Delta[/math]CXO. Thus, we know that OA = OC.[br][br]Similarly, note that Y is the midpoint of BC because [math]l_2[/math] is the perpendicular bisector of BC. From this, we know that [math]\angle[/math]BYO = 90 = [math]\angle[/math]CYO. Furthermore, we know that OY = OY and BY = CY. Therefore, by SAS congruence, we know that [math]\Delta[/math]BYO = [math]\Delta[/math]CYO. Thus, we know that OC = OB.[br][br]By transitivity, we know that because OC = OB and OA = OC, we have that OB = OA.[br][br]Since OA = OB = OC, by the definition of circle, we have that A, B, and C are all common points on a circle, E, centered at O with radius OA.[br][br]Also note, as a consequence of our result that OA = OB = OC, we know that for every point on the perpendicular bisector of each side of our triangle, [math]\Delta[/math]ABC, they are all equidistant from the endpoints of the respective side.[br][br]([math]\Longleftarrow[/math])[br][br]Let [math]\Delta[/math]ABC be a triangle in the hyperbolic plane. For any circle that contains A, B, and C, the center of the circle is equidistant from A, B, and C. Using our consequence from the previous section, we know that the center must lie on each of the perpendicular bisectors of [math]\Delta[/math]ABC. However, there is only one intersection, if it exists. Therefore, the intersection, if it exists, is the center of our circle which contains A, B, and C.[br][br]Note: The picture below is glitchy because of the custom tools, but as long the the perpendicular bisectors are concurrent, the proof stands. (Discussed with Dr. Cochran)[br]
Nine-point Circle
[br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br][br]Pink dot - circumcenter[br]Green dot - orthocenter[br][br][b][u]Important Points[/u][/b][br]Midpoints - from orthocenter to each vertex[br]Midpoints[br]Feet of the attitudes