Math 347

#14. Prove the intersection of 2 convex sets is also convex.

Proof (Stephanie Paden): Let D and E be convex sets. Let A and B be elements of , so that A and B are each in D and in E. Let C be a point on the segment AB with This would imply that , which contradicts the fact that set E (set D, respectively) was assumed convex. Therefore, the intersection of the sets is convex.

22. State and prove the angle-side-angle theorem.

Angle-Side-Angle Theorem: If two angles and the INCLUDED side of one triangle are congruent to two angles and the INCLUDED side of another triangle, then the triangles are congruent.

Proof (Martha Rojas):

Assume Postulate 15 (SAS Postulate.) Let and . If , then by SAS and we are done. Assume (note: means is not congruent to), then there exists a point G on the ray by F such that . Hence, by SAS,

. Since we assumed and since , . Therefore, . Thus, by problem # 7 and are the same line segment. Therefore, since the segment can only intersect in one point, F and G are the same point. Hence, . But this contradicts with our assumption that . Therefore,

, and so by SAS .

23) Prove that if two angles of a triangle are congruent, the sides opposite these angles are congruent.

Proof (Sharmaine Alexander):

Show that triangle BAC is congruent to triangle BCA. We are given that angle A is congruent to angle C.

We show that triangle BAC is congruent to BCA: Triangle BAC is congruent to itself in another order. Angle A is congruent to angle C, angle C is congruent to angle A, and line AC is congruent to line CA by the reflexive property. Since angle A is congruent to angle C and angle

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C is congruent to angle A and line segment is congruent to line segment , by ASA theorem, triangle BAC is congruent to triangle BCA. Therefore the corresponding side is congruent to by definition of congruence.

QED. 24). Prove an equiangular triangle is equilateral.

<AFig 1. Fig 2.

Proof (by Maria Rivas): Since <A <C, then ∆ABC is isosceles with base and equal sides . (by problem 23)Similarly, since <A <B then ∆ABC is isosceles with base and equal sides .(by problem 23)Since and , then by the transitive property . Therefore all sides are congruent. Hence an equiangular triangle is equilateral.//

25. State and Prove the theorem of side-side-side triangle congruence.

Side-side-side (SSS)If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

C

AB ED

F

Show that If , and , then the two triangles are congruent.

Proof (Erika Rey):Suppose that , then the two triangles are congruent by SAS, and we are done.

Now we will prove by contradiction that . Assume

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J

D

HF

HJ

E

F

If we draw a line ,so that .There exist a point H on the same side of the ray , such that If F and H are not the same point then and can only intersect at D. If we draw and then the by SAS. This implies that by the definition of congruent triangles. Also and

. Both and are isosceles triangles by the definition of isosceles triangles. If we take the point J to be the midpoint of , then the angle bisector of contains and the angle bisector of also contains by problem 28. This implies that D must be equal to E. This contradicts our assumption that

because D can not be equal to E. Therefore , and the triangles are congruent by SAS.

Problem #26:State and prove the theorem of Side-Angle-Angle triangle congruence.SAA: If two angles and a side of the first triangle are congruent to the corresponding parts of the second, then the triangles are congruent.

A

B

CF

E

DG"G'

Let the given SAA correspondence be: <A is congruent to <D, <C is congruent to <F, and segment AB is congruent to segment DE.

Proof (Bolivia Brandt): If segment AC is congruent to DF, then ∆ ABC is congruent to ∆ DEF by ASA and we are done.We need to prove that segment AC is congruent to segment DF. Let point G be a point on ray DF such that DG = AC.Then we have that ∆ ABC is congruent to ∆ DEG by SAS. By corresponding parts, we have <EGD is congruent to <C congruent to <EFD.

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There are two cases: either (1) DF>DG, shown as G=G’, or (2) DF<DG, shown in the diagram as G=G’’. In case (1), <EG’D is an exterior angle of ∆ EFG’ and <EG’D is congruent to <EFD, a remote interior angle of the triangle; in case (2), <EFD is an exterior angle of ∆ EFG” and <EFD is congruent to <EG’’D, a remote interior angle of the triangle. In either case, by the exterior angle theorem (38), the exterior angle must be strictly larger than either remote interior angle, and so we have a contradiction.Therefore, point F = G. Hence ∆ ABC is congruent to ∆ DEF by ASA.

27A. Is it possible for two lines to intersect insuch a way that three angles formed measure 30, 80,and 30?No: B | | A-------X-------C | | D

Given: line AB and line CD intersect at point X.Proof (Joy):If <AXB and <CXB are a linear pair, then they are supplementary. (postulate 14). By definition, <AXB and <CXB share a common ray XB, and A, X and C are collinear, then <AXB and <CXB are supplementary.

Since <AXB and <CXD are vertical angles, based on Problem16, <AXB is congruent to <CXD.

If <AXB = 30 then <CXB must equal 150.If <AXB = 30 then <CXD must equal 30.Then <BXC and <AXD must both = 150 not 80.

27B. Carefully state the conditions under which it is impossible for two lines to intersect in four angles of measures a, b, c and d.

Solution (Joy): Conditions under which the angles are possible:

a=c, d=c, a+b=180, d+c=180.

Conditions which are impossible:If any of the following inequalities holds, then it is impossible to construct two lines with the given angle measures:a+b does not equal 180d+c does not equal 180a does not equal c

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d does not equal c

28. Proof (Ryan Flores): Let E be any point on , the perpendicular bisector of given segment . Let A and B be the given endpoints of .

By the definition of perpendicular bisector, and . Also, by the Reflexive Property.

By SAS (Postulate 15) .= by definition of congruent triangles (corresponding parts), and AE=BE

(definition of congruent segments).Any point E on the perp. bisector is equidistant to the endpoints of .

29. Prove that given a point on a line in a plane, there is exactly one line perpendicular to the given line through that point.

Given: A point on a line in a plane,Prove (Keith Whitson): There is exactly one line perpendicular to the given point through a line in the plane.

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R

T

Q

W

L

J

Existence: Let R be a plane and let T, Q be points that determine a line in that plane. Let L be the given point on line TQ. Let W be a point not on line TQ to create a ray LW such that angle QLW is a right angle by the angle construction postulate. Therefore, ray LW is perpendicular to line TQ, and so is line LW.

Uniqueness: Pick another point J not on line TQ such that angle QLJ is a right angle by the angle construction postulate and so line LJ is perpendicular to line TQ. This is a contradiction to the angle construction postulate because the two angles much have the same endpoint because they have because they have the same angle measure so they must be the same angle and so the line is unique.

Therefore, given a point on a line in a plane, there is exactly one line perpendicular to the given point through a line in the plane.

32.) Ryan Flores

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Let ABC be a having side AC AB.

Prove ABC ACB.

Since AC AB, make AD AB and join BD.

Since ADB is an exterior of the BCD, it is greater than the interior and opposite DCB.

But ADB ABD, since the side AB AD.

ABD DCB=ACB. ABC ACB. In any , the opposite the greater side is greater.

Problem 33 can be obtained by emailing Prof. Jones at mjones@csudh.edu. It is a large JPEG file.

34. Prove the triangle inequality: in any triangle ABC, AB + BC is greater than or equal to AC.

Proof (by Mitsuyo Clark)Let ABC be any triangle. Construct a point D on the line AB such that A-B-D and BC = BD. Then AD = AB + BD by definition of three points. ---(1).

ACD = ACB + BCD by Angle Addition Postulate. Hence ACD > ACB and ACD > BCD. ---(2)Since the triangle BCD is isosceles with BD = BC, then BDC is congruent to BCD. --- (3) From (2) and (3), ACD > BDC by transitive property. Then by problem 33, the side AD is greater than the side AC. Hence AD > AC in terms of distance. ---(4)From (1) and (4), AB + BD = AD > AC, and BD = BC, by transitive property, we have AB + BC > AC.

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C DA

B

35. Prove that if two lines in a plane are both perpendicular to the same line, then they are parallel.

A l

B m

nGiven: Line l is perpendicular to line n at point A.

Line l is perpendicular to line n at point B.

Prove: Line l is parallel to line m.

Proof (Joel Negrete): We will assume that the two lines (l and m) are not parallel and will intersect at a point P.

l A

B P m n

If the two lines intersect then we have two perpendicular lines to line n through point P which is a contradiction (proved in problem 29).

Therefore lines l and m are parallel.

36. Prove that there is at least one line through a given point not on a given line that is parallel to the given line.

P n

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l

m

Given: A line l and a point P not contained in line l

Prove: we can construct a parallel line to line l through point P.

Proof (Alejandro): Given a line and a point not on the given line we can find a plane that contains both the line and the point. By Postulate 7: There is at least one plane containing any three given points. If the points are non-collinear, then there is only one such plane.

Since both the line l and the point P are on the plane S, we can construct a perpendicular line m through point P to line l. By problem # 31 - given a point not on a given line in a plane, there is exactly one line perpendicular to the given line.

Since point P is contained in line m and line m is contained in the same plane as point P and line l, we can construct a line n perpendicular to line m through point P which is contained in line mBy problem # 29 - given a point on a line in a plane, there is exactly one line perpendicular to the given line through that point.

Since both line l and line n are perpendicular to the same line m, then line l is parallel to line n. By problem 35, if two lines in a plane are both perpendicular to the same line, then they are parallel.

Homework Problem # 37a) Define alternate interior angles.

When two lines are crossed by a transversal, two angles located inside two lines and on opposite sides of the transversal are called alternate interior angles.

b) Prove that if two lines are cut by a transversal, and if one pair of alternate interior angles is congruent, then the other pair of alternate interior angles is also congruent.

Given: m1 m3Prove: m2 m4 (Susana Granados)

Statements/Conclusion Justification 1) m1 m3 1) Given2) m1 + m2 = 180 2) Supplementary angles. m3 + m4 = 180 Supplementary angles.3) m1 + m2 = m3 + m4 3) Set equal to each other.4) m1 + m2 = m1 + m4 4) Given, Substitute m1 with m3

4 3

21

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5) m1 + m2 = m1 + m4 5) Subtract m1 from both sides - m 1 - m 1 6) m2 = m4

Problem # 40 presented by Juan CastilloTwo nonadjacent angles made by the crossing of 2 lines by a third line, one angle being interior, the other exterior, and both being on the same side of the line.

1 2 l3 4

5 6 m7 8

tIt is given that angle 2 is congruent to angle 6. We want to prove that angle 1 is congruent to angle 5, angle 7 is congruent to angle 3, and angle 8 is congruent to angle 4.Proof:By the problem on vertical angles (16(b)), angle 2 is congruent to angle 3 and angle 6 is congruent to angle 7. Therefore, angle 3 is congruent to angle 7 by substitution or transitive property. By definition of supplementary angles: the sum of the measurements of angle 2 and angle 1 is 180 degrees. As a result, the measurement of angle 2 is equal to 180 degrees minus the measurement of angle 1. Similarly, the sum of the measurements of angle 6 and angle 5 is equal to 180 degrees. As a result, the measurement of angle 6 is equal to 180 degrees minus the measurement of angle 5. Since angle 2 is congruent to angle 6, then 180º - m<1 = 180º - m<5.Therefore, the measurement of angle 1 is equal to the measurement of angle 5, which implies that angle 1 is congruent to angle 5. By definition of vertical angles angle 1 is congruent to angle 4 and angle 5 is congruent to angle 8. Since angle 1 is congruent to angle 5, then angle 8 is congruent to angle 4 by the transitive property.

41. Prove that if two parallel lines are cut by a transversal, then alternate interior angles are congruent.

Proof (Stephanie Paden): Below is my proof for #41. It was determined by the class and Dr. Jones that my reason for statement #2 could not be ‘by problem #40’, since we are not given a set of corresponding angles.

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1. a || b; t is transversal 1. Given 2. 2. By Problem #46 & #48. (corr s)

3. 3. Vertical 4. 4. Transitive Property

For reason #2, problem #48 asks for angles created when 2 parallel lines crossed by a third line.Prof. JONES’ note: #46 is not a valid reason for line 2. #48 is a valid reason, but then you need to prove some other set of angles, such as corresponding angles, is congruent. This is a long way to get to #41. To prove #41 without using #48, use proof by contradiction: Let the intersection of the transversal with line b be point B. If angles 2 and 6 are not congruent, then there is a line b’ with angle 6’ at B congruent to angle 2. Then by #39, b’ is also parallel to a. But by postulate 16, there can only be one line through B parallel to a. This is a contradiction. Thus the alternate interior angles 2 and 6 are congruent, and by 37b, angles 1 and 5 are also congruent.//

Problem 42Let the line M be the transversal to the lines L and NAlso L // N

M M

N P K P 1 2

L 3 L 3

Proof (Brahim Karam):(1 ) Then by theorem 41, we have m< 1 = m<3 (alternate interior angles) Now, let’s assume that there exist another line K through P such that: K // L and m< 2 ≠ m< 1

(2 ) Then, by theorem 41, we have m< 2 = m< 3 (alternate interior angles)(3) therefore, by (1) and (2) we have m< 2 = m< 1 (transitivity) By statement (3) is a contradiction to the statement m< 2 ≠ m< 1

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Therefore, there is only one line that is parallel to L through the point P

43. A close translation of the Parallel Postulate as Euclid gave it is: If a third line crosses two other lines, and the interior angles on the same side of the third line have an angle sum less than 180 degrees, then the lines, if extended sufficiently, meet on the side in which the angle sum is less than 180 degrees.

a. Restate the postulate in your own words. b. Prove Euclid’s version as a consequence of Postulate 16.

Solution (Ryan Flores, with substantial revision by Prof. Jones):

a) Postulate 16 (Parallel Postulate) – If given 3 non-collinear points ABC, and if a line passes through points Aand B, then there exists a unique line through point C parallel to the given line through A and B.b) Let m be a line and let o be another line with transversal line p, and with a same-side interior angle sum of less than 180. Let o intersect p at point A. We need to show that o intersects m on the side on which the angle sum is less than 180. Construct a transversal line n perpendicular to line m and passing through A (possible by 31), and construct a line l perpendicular to n and passing through A (possible by 29). By problem 36, line l is parallel to line m. Because we know that l || m, we can say that m1 = 90, m2 = 90, m3 = 90, and m4 = 90. We know this by the results of problems 41 and 16. Therefore, m1 + m3 = 180 and m2 + m4 = 180.

Now, line o also passes through point A. By Post. 16, we know there cannot be more than one line || to m through point A. Knowing this, we can say that if o WERE || to m through A then o and l must be the same line. Thus o is not parallel to m and the lines meet.

o

p m

l

A

B

It remains to show that o and m meet on the side in which the same-side interior angle sum is less than 180.Since l and m are parallel, and p is a transversal, we know that the same-side interior angle sum is equal to 180 on both sides (by 41 and linear pairs). Let B be a point on line o on the side of p in which the sum is greater than 180. By plane separation, and angle addition, ray AB is on the opposite half-plane from line m, and hence does not intersect line m. Thus the lines must meet on the side in which the sum is less than 180.

Problem 44A-Let the line M be the transversal to the lines L and N. Also L // N

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M Q

N P 4 11’

L 3 2

Proof (Brahim Karam):It follows by Theorem 43 that <1 and <2 are supplementary, as are <3 and <4. Let Q be another line // to L through the point P, such that

First case: m< 1’ is less that m<1.Then, the sum of m< 1’ and m< 2 is less than 180 degrees. Also, <1’ and <2 are on the same side of the transversal M Therefore, by theorem 43 the line Q and the line L are not parallel, which contradicts the given that Q //LSecond case: m<1’ is bigger than m<1Then the sum of m<3 and m<4 is less than 180 degrees Hence, again by theorem 43 Q is not parallel to L

In both cases, we have if a line through the point P and parallel to the line L then it is unique.

B- (Brahim)Yes, theorem 43, stated as, “If a third line crosses two other lines, and the interior angles on the same side of the third line have an angle sum less than 180 degrees, then the lines, if extended sufficiently, meet on the side in which the angle sum is less than 180 degrees,” and which is in the form ┐Q┐P, is equivalent to the statement, “If two parallel lines are cut by a transversal, interior angles on the same side of the transversal are supplementary,” which is in the form of P Q. These two forms are logically equivalent and one is the contrapositive of the other.

45.

A C

BD E

Prove: m<A +m<ABC+m<C = 180 degrees (Edgar E. Perez)

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1.) Line DE ll line AC by construction2.) m<DBE = 180 degrees by linear pair supplementary definition3.) m<DBA + m<ABC+m<CBE = 180 degrees by linear pair/supplementary4.) m<A = m<DBA By #47 alternate interior angles of parallel lines are congruent

m<C = m<CBE

Hence, m<A+m<ABC+m<C = 180 degrees by transitivity

49. Show that the image of a segment under an isometry is also a segment.

B

A

C C'

A'

B'

Proof (Martha Rojas):

We know that a transformation T is an isometry if it preserves distance.

Let T be any isometry and be any segment with T(A) = A’ and T(B) = B’. Because T is an isometry, AB = A’B’. Therefore, we only need to show that , that is, show that a point C is on if and only its image T(C) = C’ is a point on . Since we know T(A) = A’ and T(B) = B’, we only need to consider points C and C’ between those points. To show

, we need to show two things: 1) and 2) .

To show that , suppose C is on and between A and B. Then, AC + CB = AB by definition of between. Since T preserves distance, A’C’ = AC and C’B’ = CB. Therefore, A’C’ + C’B’ = A’B’. This implies that C’ is between A’ and B’ and thus C’ is on .

To show that , we do the opposite. Suppose C’ is on and between A’ and B’. Then, A’C’ + C’B’ = A’B’. Again, since T preserves distance, it must be that its preimage C satisfies AC + CB = AB. Therefore, C is between A and B and on .

Therefore, or the image of a segment under an isometry is also a segment.

PROBLEM #50Show that the image of a ray under an isometry is also a ray.

Proof (Bolivia Brandt): From the definition of isometry and from 49, we know: Every isometry preserves angle measure, betweennes, collinearity, and distance (length of segments).Suppose we are given ray XY, and that the images of points X and Y are X’ and Y’.

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A

B'

X X'

B

A'

YY'

By problem #49: A segment under an isometry is also a segment. A ray contains an infinite amount of segments and each segment is preserved under the isometry; in other words, for each segment XZ of ray XY, segment X’Z’ is a corresponding and congruent segment of ray X’Y’.Therefore, we can say that any other point in the ray XY will be on the ray X’Y’ according to the above definition. Hence, the image of a ray under an isometry is also a ray.

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51. Show that the image of an angle under an isometry is also an angle.

Given: PQRProve: P’Q’R’

Proof: (Erika Rey)

Q

P

R

Q'P'

R'

is isometric/congruent to and is isometric/congruent to by problem 50. Since both and share Q’, by angle construction two nonlinear rays with a common vertex form an angle. Therefore the image of an angle under an isometry is also an angle. QED

52. Show that segments congruent under the new definition are congruent under the old definition.Solution (Keith Whitson):Old Definition: Two segments are congruent if the have the same length.New Definition: Two objects in a plane are said to be congruent if there is an isometry with T(A)=B.Isometry: A transformation is an isometry if it preserves distance.

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By #49, the image of a segment is also a segment. Let T be an isometry and segment AB be any segment with T(A)=A’ and T(B)=B’. Because AB=A’B’ under the new definition, which preserves distance, segment AB is congruent to segment A’B’ under the old definition because the two line segments have the same length.

53) Show that congruent angles under the new definition are congruent under the old definition.

*OLD DEFENITION - Two angles are congruent if their measures are equal.*NEW DEFENITION – Two objects A and B in a plane are said to be an isometry with T(A) = B.

***Show that T(< PQR) is congruent to (< P’Q’R’)****Proof (Belinda Vargas):Assume P’, Q’, and R’ are the images of P, Q, and R, respectfully, under the isometry. Since isometries preserve distance, we know PQ=P’Q’, PR=P’R’, and QR=Q’R’.

Therefore, triangle PQR is congruent to P’Q’R’ (by SSS). So, by corresponding parts < PQR is congruent to < P’Q’R’ (by old definition)

R'

P'

Q'

R

P

Q

54. Assume that two triangles are congruent under the new definition. Then there is an isometry which maps one triangle onto the other. Show that these triangles are congruent under the old definition.Proof (by Mitsuyo Clark)

Given an isometry T with T(ΔABC) = ΔA'B'C' and ΔABC ΔA'B'C' under the new definition. Since distance is preserved under an isometry T, AB = A'B', BC = B'C', and AC = A'C'. By problem 52 (segments congruent under the new definition are congruent under the old definition), , , and . By SSS theorem, ΔABC ΔA'B'C' under the old definition.

Problem 55 can be obtained by emailing Prof. Jones at mjones@csudh.edu. It contains scanned images in a word document.

56). Prove that rotations are isometries.

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Given: A’ and B’ are the rotation images of A and B after a rotation about C. Prove: Proof: (Maria Rivas): By cases.Case 1. A=C

Since A=C, AB=CB. By definition of rotation CB=CB’. By transitivity, AB=A’B’. (AB=CB=CB’=A’B’). Similarly if B=C.Case 2. ABC are collinear but distinct.

By segment addition, AB= AC+CB and A’B’= A’C+CB’. By definition of rotation CB= CB’ and CA=CA’. By transitivity, AB=A’B’.Case 3. A, B ≠ C and noncollinear

By definition of rotation, CA=CA’, CB=CB’, and ACA’ .

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By definition of congruence, m hence m + m = m ACA’ and m 2+ m 3= m BCB’. By transitivity, m + m = m 2+ m 3, it follows that, m = m 3. By SAS theorem, ∆ABC ∆A’CB’. Therefore, by corresponding parts of congruent triangles, AB=A’B’.

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#57) Prove that translations are isometrics.

Proof (Sarah O’Crowley): We must show that the distance between two given points J and K is preserved under translation.Case 1 : If one pair of opposite of a quadrilateral is parallel and congruent, then the quadrilateral is a parallelogram. JJ’ and KK’ are parallel and congruent, so JJ’KK’ is a parallelogram. Since the opposite sides of a parallelogram are congruent, JK is congruent to J’K’ and JK = J’K’.

J J’

K K’

Case 2: When JK is moved along a vertical line It follows from the definition of translation that JJ’=KK’. 1) K’K = K’J’ + J’K 2) JJ’ = J’K + KJ

By substitution and cancellation, K’J’ =KJ.

K’ *

*J’

*K

*J

60. Prove that the composition of two isometries is also an isometry. Note: A transformation T is an isometry if it preserves distance. Proof (Sharmaine Alexander): Let F and G be two isometries. That means F and G both preserve distance. For any two given points, p and q, we have the following:G (p) = p’ and G (q) = q’. Since G is an isometry we can say that pq = p’q’. Also F (p’) = p” and F (q’) = q”. Since F is an isomtery we can say that p’q’ = p”q”. Therefore pq = p’q’ = p”q”. Thus F ○ G (p) = p” and F ○ G (q) = q”. Therefore the composition of F and G is an isometry.

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61. Hint: Showing congruence under the new definition means that one must describe an isometry, or a sequence of isometries (allowed because of 54), in which the image of the first object is the other object.

a. Prove that two segments congruent under the old definition are congruent under the new definition.

Given: Segment AB is congruent to segment CD (old definition).Proof (Keith Whitson): Show that there is a distance-preserving transformation that maps segment AB onto segment CD. Let m be the perpendicular bisector of segment AC. Then the reflection rm maps segment AB onto a segment A’B’, and A’=C by construction. Let n be the bisector of angle B’CD. Because CD=AB=A’B’=CB’ and n is the bisector, rn (segment CB’) = segment CD. Thus, rn○rm (segment AB) = segment CD, and so segment AB is congruent to segment CD. QED.

n

m

A

B D

B'

C = A'

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b. Prove that two angles congruent under the old definition are congruent under the new definition.

Given: <ABC is congruent to <DEF (old definition)Prove: Show that there is a distance-preserving transformation that maps <ABC onto <DEF. Proof: Let t be the translation that connects vertex B to vertex E, then let r be the rotation by the angle A’B’D that rotates rays A’B’ and B’C’ into the positions of DE and EF respectively (it may be necessary to reflect across the bisector of <DEF to get proper alignment of the points D, E, and F with A, B, and C). Thus, r composed with t (and perhaps a reflection) is an isometry that preserves the angle measurement of the angles. Therefore, two angles congruent under the old definition are congruent under the new definition.

C''=F

A''=D

C'

A'

A

B

C

E=B'

62). Prove that two triangles congruent under the old definition are congruent under the new definition.

Old definition: Two triangles are congruent if there is a correspondence between the angles and the segments of each triangle such that the corresponding angles and segments are congruent. New definition: Two objects in the plane are said to be congruent (A congruent to B) If there is an isometry T (A) =B.

Prove that ∆ABC ∆ A’B’C’ (which are congruent under the old definition) are also congruent under the new definition. In other words, there is an isometry U such that U (∆ABC) = (∆ A’B’C’).

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Proof (Maria Rivas): Given two congruent triangles, wlog we can assume ∆ABC ∆ A’B’C’ by SSS theorem. (Old definition). Therefore, AB = A’B’, BC= B’C’ and AC= A’C’.Since AB =A’B’, there exists an isometry T such that T (AB) = A’B’.

Case 1). If T(C) = C’, then set U=T and the proof is done because every isometry preserves distance.

Case 2). Because line A’B’ divides the plane in half and because T preserves angles (by 61b), there are only two possibilities for the isometry: either T(C) = C’ or

C’ (which can also be expressed as T(C) = C”). Since T (C) = C’ was proven in case 1, wma that the isometry U= .Therefore U (∆ABC) = (∆ A’B’C’) and so ∆ABC ∆ A’B’C’ under new definition.

Problem 63 (Susana Granados): The Pythagorean Theorem states that if ABC is a right triangle with legs of length a and b, and a hypotenuse of length c, then a2 + b2 = c2. Prove the Pythagorean Theorem.

We start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively.

Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.

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The square has a square hole with the side (a-b). Summing up its area (a-b)2

and 2ab, the area of the four triangles (4·ab/2), we get

c2 = (a - b)2 + 4 * (ab/2)

c2 = (a - b)2 + 2ab

c2 = a2 – 2ab + b2 + 2ab

c2 = a2 + b2. QED

64. Show  that  with  SSA, there  can  be  two  different, incongruent  triangles  which  bothsatisfy SSA.

Solution (Ventress Sulcer):SSA -  If  two  sides  and  a  non - included  angle  of  one  triangle are  congruent  to  the  corresponding  parts  of  another  triangle ,then  the  triangles  are congruent. 

 Given  two  triangles  ABC and  XYZThe  line  AB  is  congruent  to  line  XY, line  AC  ls  congruent to  line  XZ  and.angle C is  congruent  with  angle  Z .(SSA)  Even  though  two  sides  and  two  angles  are congruenttriangles ABC  and  XYZ  are  not  congruent.

66. What happens to the plane (or objects in the plane, if you prefer) if you perform two reflections? Put another way, what is the result of performing 2 reflections across lines l1 and l2 in the plane? There are three possible cases for the two lines of reflection: l1 and l2 are identical, they are parallel to each other, or they intersect.

Prove that if the lines are parallel, the result is a translation.

Proof (Alejandro Vidal): If line m and line n are parallel, then rn ο rm is the translation with direction perpendicular to the lines m and n, and with magnitude twice the distance between m and n, in the direction from m to n.Case 1 Case 2 A P A’ x x

Math 347

x x A A’y Q y

y 2x + y A”

m n m n A”

Proof Suppose m and n are parallel lines. Let A’ = rm (A) and let A” = rn (A’) = rn ο rm (A). Since segment AA’ and segment A’A” are perpendicular to parallel lines, the three points A, A’, and A” are collinear. So line AA’’ is perpendicular to line m.

To find the length of AA”, notice that lines m and n are perpendicular bisectors of segment AA’ and segment A’A”. Consequently, if the distance between lines m and n is x + y as show above, then AA” = 2x + 2y.

rn ο rm = TAA” AA” = 2(PQ) Segment AA” is perpendicular to lines m and n. Thus the composition of the two reflections is the translation TAA’’, where the translation is perpendicular to lines m and n in the direction from m to n, and is by a distance twice that of m to n.

Note: rn ο rm ≠ rm ο rn .