Proof Portfolio (to turn in)
Table of Contents
- Chapter 7
- Proof Theorem 7.7
- Proof Lemma 7.1
- Proof Theorem 7.2
- Chapter 2
- Betweenness Proposition I
- Rhombus proof
- Chapter 3
- Proof 3.3.19
- Isosceles triangle proof of congruent base angles
- Ceva's theorem
- Chapter 4
- Proof 4.3.11
- 4.3.14: Proof of Ptolemy's Theorem
- Exercise 4.3.6
- Proof 4.3.23
- Chapter 5
- Proof of np center
- Chapter 6
- Proof 6.3.4
- Proof 6.3.8
- Chapter 8/9
- Proof 8.3.3,4,5
- Proof 8.3.34
- Proof 8.3.17
- Chapter 11
- Proof 11.10
- Copy of Exercise 11.3.3
- Proof 11.5.12 (circumcircle proof)
- Animation of triangle congruence
- Triangle Congruence
Proof Theorem 7.7
In a projective plane, any two distinct lines have exactly one point in common.
Proof:[br]Consider two distinct lines [i]l[/i] and [i]l'. [/i]Projective Axiom 3 tells us that [i]l [/i]and [i]l' [/i]have at least one point, Q, in common. Now suppose there exists a second point, P, distinct from point Q and shared by [i]l[/i] and [i]l'. [/i]That is, we have two distinct points, P and Q, that have two lines, [i]l [/i]and [i]l', [/i]in common. This contradicts Projective Axiom 2. Thus, [i]l [/i]and [i]l'[/i] cannot have a second point in common. Therefore, [i]l[/i] and [i]l' [/i]only have one point in common. We can then conclude that any two distinct lines in a projective plane have exactly one point in common.
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 3.3.19
Proof of hypotenuse-leg criterion for congruence
Consider two right triangles ABC and DEF with right angles A and D, AB congruent to DE, and BC congruent to EF. [br][br]Construct a point X on the ray starting at C and going through A, such that AX is congruent to DF. Then connect X to B to form right triangle ABX. [br][br]Then we can see that AB is congruent to itself, angle CAB is congruent to angle XAB (since they are both right angles), and line XB is congruent to line BC (since XF is congruent to EF and EF is congruent to BC).[br][br]But since we know that ABC and ABX are right triangles, we know that angle ABC and angle ABX are congruent, else AB would no longer be perpendicular to XAC and we would no longer have two right triangles. Then by SAS, we can see that triangles ABX and ABC are congruent.[br][br]Now since XB is congruent to BC, we know that triangle XBC is isosceles by Definition 20. By Euclid's fifth proposition, we know that the base angles, angle BCX and angle BXC, are congruent. Then we can see that angles X, C, and F are congruent since we defined AX to be congruent to DF. [br][br]Then, by replacing X with F, B with E, and A with D in triangle ABX, we can easily see that AB is congruent to DE, angle CAB is congruent to angle FDE, FE is congruent to BC, DF is congruent to AC, angle ABC is congruent to angle DEF, and angle C is congruent to angle F. Therefore, it is clear that two right triangles with a congruent leg and hypotenuse are congruent.
Proof 4.3.11
Cyclic quadrilateral ABCD.
Conjecture: The perpendicular bisectors of a quadrilateral are concurrent if and only if the quadrilateral is cyclic.
[b]Forwards implication: [/b]If the perpendicular bisectors of a quadrilateral are concurrent, then the quadrilateral is cyclic. [br][b]Proof of forwards implication (by direct proof): [/b]Consider the quadrilateral ABCD. Assume that the perpendicular bisectors of ABCD are concurrent. That is, the perpendicular bisectors of ABCD all intersect at some point, P. [br]Consider the side AB of the quadrilateral. Because any point on a perpendicular bisector of a line segment must be an equal distance from each endpoint (proved in class), we can assume that [math]AP\cong BP[/math]. [br]Similarly, when considering side BC, we can assume that [math]BP\cong CP[/math]. [br]When considering side CD, we can assume that [math]CP\cong DP[/math]. [br]Finally, when considering side DA, we can assume that [math]DP\cong AP[/math]. [br]That is, [math]AP\cong BP\cong CP\cong DP[/math]. Thus, the points, A, B, C, D lie on a circle centered at P. Therefore, the quadrilateral ABCD is cyclic. [br][br][b]Backwards implication: [/b]If a quadrilateral is cyclic, then the perpendicular bisectors of the quadrilateral are concurrent. [br][b]Proof of backwards implication (by direct proof): [/b]Consider the cyclic quadrilateral ABCD.By definition, we know that the points A, B, C, D must lie on a circle such that A, B, C, D are all equal lengths from the center of the circle, P. That is, [math]AP\cong BP\cong CP\cong DP[/math]. [br]Given that a point lies on a perpendicular bisector of a segment if it is an equal distance from both endpoints (proved in class) and [math]AP\cong BP[/math], we can conclude that P is on the perpendicular bisector of side AB. [br]Similarly, because [math]BP\cong CP[/math], we can conclude that P is on the perpendicular bisector of side BC. Because [math]CP\cong DP[/math], we can conclude that P is on the perpendicular bisector of side CD. Finally, because [math]DP\cong AP[/math], we can conclude that P is on the perpendicular bisector of side DA. [br]Because P lies on every perpendicular bisector of quadrilateral ABCD, we know that the perpendicular bisectors are concurrent at point P.
Proof of np center
Nine Point Circle with Euler Line
Use coordinates to prove that the nine-point center is the midpoint of the segment from the orthocenter to the circumcenter.
[math][/math](NOTE: Assume all points correspond to Fig. 5.14 on page 131 of the textbook- I tried to rename the points but every time I hover over a point or line it pops up with the name and coordinate and doesn't let me right click)[b][br]Proof:[br][/b]Create [math]\triangle ABC[/math]such that AC lies on the x-axis and B lies on the positive y-axis. That is, A=(-a,0), B=(0,b), and C=(c,0). Consider the nine point circle of the triangle. Let Q be the midpoint of AB, R be the midpoint of BC, and P the midpoint of AC. Let the foot of AB be point D, the foot of BC be point E, and the foot of AC be point F (the origin). Allow H to be the orthocenter and notice that it lies on the y-axis. Let S be the midpoint of HA, T be the midpoint of HB, and U be the midpoint of HC. Furthermore, call the center of the nine point circle N and the circumcenter of [math]\triangle ABC[/math]O.[br][br]To begin,we will find the coordinates of the orthocenter, H. Notice, H lies on the y-axis, thus the coordinates of H are (0,h). Notice, by definition of the orthocenter, the line AE goes through H. So we can determine the value of h by solving for the equation of the line AE. [br]AE is perpendicular to BC, thus[br](slope of AE)(slope of BC)=-1. [br][math]slopeofBC=-\frac{b}{c}[/math]. [br]So [math]slopeofAE=\frac{c}{b}[/math]. [br]The equation of AE is thus [br][math]y=\frac{c}{b}x+h[/math]. [br]If we plug in the point A=(-a,0) and solve for h, we find that[br][math]h=\frac{ac}{b}[/math]. [br]So the orthocenter, H, is located at the point [math]\left(0,\frac{ac}{b}\right)[/math].[br][br]Next, we will find the coordinates of the circumcenter, O. To do this, we will find the intersection of two of the perpendicular bisectors [math]\triangle ABC[/math]. The midpoint of AC occurs at point P. We can easily determine the coordinates of P using the midpoint formula to be [math]P=\left(\frac{c-a}{2},0\right)[/math]. Because P lies on the x-axis, we know that the equation of the line perpendicular to the x-axis, through P, is [math]x=\frac{c-a}{2}[/math]. [br]Next, we will find the equation of the perpendicular bisector of BC. The midpoint of BC is R. Using the midpoint formula we know that [math]R=\left(\frac{c}{2},\frac{b}{2}\right)[/math] .The slope of this line is going to be perpendicular to the slope of BC. So as before, [math]slope=\frac{c}{b}[/math]. By substituting the point R, into the slope-intercept form of the equation of the perpendicular bisector, RO. We can determine that equation to be [math]y=\frac{c}{b}x+\frac{b^2-c^2}{2b}[/math]. By substituting [math]x=\frac{c-a}{2}[/math] into this equation, we can find the y value of the circumcenter. We can then determine the coordinates of the circumcenter to be [math]O=\left(\frac{c-a}{2},\frac{b^2-ac}{2b}\right)[/math]. [br]Finally, we can determine the midpoint of the orthocenter, H, and the circumcenter, O, using the midpoint formula. After some simplification, this becomes: [br][math]midpoint_{OH}=\left(\frac{c-a}{4},\frac{b^2+ac}{4b}\right)[/math]. [br][br]Now, we can determine the center of the nine point circle, N, using the two chords FT and FP. We must first determine the coordinates of T. T is the midpoint of HB, thus [math]T=\left(0,\frac{b^2+ac}{2b}\right)[/math]. Therefore, the midpoint of the chord FT is [math]midpoint_{FT}=\left(0,\frac{b^2+ac}{4b}\right)[/math]. The midpoint of the chord FP is [math]midpoint_{FP}=\left(\frac{c-a}{4},0\right)[/math]. Because FT lies on the y-axis and FP lies on the x-axis, we can easily determine the center of the nine point circle, N, to be [math]N=\left(\frac{c-a}{4},\frac{b^2+ac}{4b}\right)=midpoint_{OH}[/math]. [br]Thus, we see that the nine-point center is the midpoint of the segment from the orthocenter to the circumcenter.
Proof 6.3.4
[b]Let [math]g\left(P,Q\right)=[/math][math]g\left(P,Q\right)=max\left(x_P,x_Q\right)+max\left(y_P,y_Q\right)[/math]. [br][br]a)[/b] We can show that [math]g\left(P,Q\right)[/math] is not a metric. Consider the points [math]P=\left(-2,0\right)[/math] and [math]Q=\left(-1,-1\right)[/math]. Then, [math]g\left(P,Q\right)=max\left(-2,-1\right)+max\left(0,-1\right)=-1+0=-1.[/math] This contradicts the first metric axiom, thus [math]g\left(P,Q\right)[/math] is not a metric. [br][b]b) [/b]does not apply [br][b]c) [/b]Does not apply
Proof 8.3.3,4,5
Consider the functions, [math]f_1\left(x,y\right)=\left(x,\frac{1}{y}\right)[/math] and [math]f_3\left(x,y\right)=\left(x+2y,y\right)[/math]. [br][br]We can show that [math]f_1[/math] is not an onto function because there are points in the plane that cannot be obtained as outputs of [math]f_1[/math]. Counterexample: The point (1,0) exists in the target but there does not exist a corresponding (x,y) in the domain. That is, there is no point (x,y) such that f(x,y)=(1,0). [br]We can show that [math]f_3[/math] is an onto function. Let [math]f_3\left(x,y\right)=\left(a,b\right)[/math]. Then [math]a=x+2y[/math] and [math]b=y[/math]. Both a and b are defined for all values of x and y. Thus, for any (a,b) in the target, there exists an (x,y) in the domain such that f(x,y)=(a,b). [br][br]We can easily show that [math]f_1[/math] is one-to-one. Consider the element [math]Y\in f_1[/math] such that [math]Y=\left(x_1,\frac{1}{y_1}\right)[/math]. We see that the x-coordinate always maps to itself, thus there is a one-to-one relationship for the x-coordinates. By rearranging, we see that [math]y_1=\frac{1}{y}[/math]. Taking the reciprocal of any real number gives another unique real number and thus the mapping between the y-coordinates is also one-to-one. That is, [math]f_1[/math] is one-to-one. [br]We can easily that [math]f_3[/math] is also one-to-one. Consider the element [math]Y\in f_3[/math] such that [math]Y=\left(x_1+2y_1,y_1\right)[/math], The y-coordinate always maps to itself, thus there is a one-to-one relationship for the y-coordinates. By rearranging, we see that [math]x_1=x-2y_1[/math]. This equation gives a unique real. Thus the mapping between the x-coordinates is one-to-one. Therefore, [math]f_3[/math] is one-to-one. [br]We can see that neither function is distance preserving. Consider the points A=(a,b) and B=(c,d). Then [math]d\left(A,B\right)=\sqrt{\left(a-c\right)^2+\left(b-d\right)^2}[/math]. For [math]f_1[/math], [math]d\left(f_1\left(A\right),f_1\left(B\right)\right)=\sqrt{\left(a-c\right)^2+\left(\frac{1}{b}-\frac{1}{d}\right)^2}\ne d\left(A,B\right)[/math]. [br]For [math]f_3[/math], [math]d\left(f_3\left(A\right),f_3\left(B\right)\right)=\sqrt{\left(a+2b-c-2d\right)^2+\left(b-d\right)^2}\ne d\left(A,B\right)[/math].
Proof 11.10
In the hyperbolic plane, the sum of the angles in any triangle is less than 180 degrees.
[b]a) [/b]Suppose that ABC is a triangle in the hyperbolic plane. Let D be the midpoint of the segment AB and let E be the midpoint of the segment AC. Construct segments AX, BX, and CZ that are perpendicular to the line DE, with the points X,Y,Z lying on the line DE. Consider the case when point X is exterior to the segment YZ as shown in the top figure (and on page 279 of textbook). [br]Note, this figure shows A to the left of YZ. Without loss of generality, if A was to the right, our argument would still hold. [br]Through AAS triangle congruence, we see that:[br][math]\triangle ADX\cong\triangle BDY[/math] and [math]\triangle AXE\cong\triangle CZE[/math]. [br]From the first triangle congruence, we know that [math]AX\cong BY[/math]. From the second, we know that [math]AX\cong CZ[/math]. Thus, [math]CZ\cong BY[/math]. Thus, we can see that quadrilateral ZYBC is a Saccheri quadrilateral because we defined the angles at Z and Y to be perpendiculars and [math]CZ\cong BY[/math]. [br]Then[br][math]\angle A+\angle B+\angle C=\angle CAB+\angle ABC+\angle BCA[br][/math][br] [math]=\angle XAD-\angle EAX+\angle DBC+\angle BCZ+\angle ZCE[/math][br] [math]=\angle YBD+\angle DBC+\angle BCZ[/math][br] [math]=\angle YBC+\angle BCZ[/math][br] [math]=2\cdot\angle BCZ[/math]. [br]Because quadrilateral ZYBC is a Saccheri quadrilateral, [math]\angle BCZ[/math] is acute. Thus, [math]\angle A+\angle B+\angle C=2\cdot\angle BCZ<2\cdot90^\circ<180^\circ[/math]. [br][br][b]b)[/b] Suppose that ABC is a triangle in the hyperbolic plane. Let D be the midpoint of the segment AB and let E be the midpoint of the segment AC. Construct segments AX, BX, and CZ that are perpendicular to the line DE, with the points X,Y,Z lying on the line DE. Consider the case when point X equals point Z (and thus also equals point E). Then we see that [math]\triangle ADE\cong\triangle CDE\cong\triangle BDY[/math]. Thus, the quadrilateral EYBC (or ZYBC) is a Saccheri quadrilateral because the angles at E and Y are right angles and [math]EC\cong YB[/math]. [br]Then[br][math]\angle A+\angle B+\angle C=\angle CAB+\angle ABC+\angle BCA[/math][br] [math]=\angle YBD+\angle YBC-\angle YBD+\angle BCD+\angle DCE[/math][br] [math]=\angle YBC-\angle ECB[/math][br] [math]=0<180^\circ[/math].