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Chapter 4-One Way to Go: Euclidean Geometry of the Plane

Chapter 4-One Way to Go: Euclidean Geometry of the Plane. Neutral Geometry: all theorems derived without making any assumptions about parallel lines Geometers were trying to extend neutral geometry to include all of Euclidean but none ever succeeded

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Chapter 4-One Way to Go: Euclidean Geometry of the Plane

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  1. Chapter 4-One Way to Go: Euclidean Geometry of the Plane • Neutral Geometry: all theorems derived without making any assumptions about parallel lines • Geometers were trying to extend neutral geometry to include all of Euclidean but none ever succeeded • Because several people were able to show Euclid’s 5th postulate is independent

  2. 4.2 The Parallel Postulate and Some Implications • Assume Postulate 16 (or Playfair’s Postulate) • Through a given point there is at most one line parallel to a given line. • Alternate def: Two lines are parallel if they are coplanar and do not intersect. • Now that we have assume the parallel postulate, we can add theorems we mentioned before (as being equivalent to the EPP)

  3. Theorem 4.2.1: If two parallel lines are crossed by a transversal, the alternate interior angles are congruent • Theorem 4.2.2: The sum of the measures of the interior angles of a triangle is 180 • Read proof on page 129 • Cor 4.2.3: The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles.

  4. Proof: • Let ΔABC exist such that D is a point on AC such that A-C-D • Know that sΔABC=180, so m<1+m<2+m<3=180 • By def of linear pair, m<2+m<4=180 • By substitution, m<1+180-m<4+m<3=180 • So m<1-m<4+m<3=0 • Or m<1+m<3=m<4.

  5. More results of EPP • Parallelogram: A quadrilateral is a parallelogram if and only if both pairs of opposite sides are parallel • Theorem 4.2.4: The opposite sides of a parallelogram are congruent • Read proof, page 130

  6. More results • Theorem 4.2.5: If a transversal intersects three parallel lines in such a way as to make congruent segments between the parallels, then every transversal intersecting these parallel lines will do likewise. • See picture and read Proof page 131 • Cor 4.2.6: If a transversal crosses three or more parallel lines in such a way as to result in congruent segments between the parallels, then every transversal will do likewise.

  7. Induction • Proof by induction: • Need to show true for n=1 • Assume true for n=k • Then show true for n=k+1

  8. Proof of Cor 4.2.6 • Shown true for n=3 (theorem 4.2.5) • Assume true for n=k, which implies • A1A2 = A2A3 = … = Ak-1Ak and • B1B2 = B2B3 = … = Bk-1Bk • Now show true for line lk+1 • Well if A1A2 = A2A3 = … = Ak-1Ak = AkAk+1 then B1B2 = B2B3 = … = Bk-1Bk = BkBk+1 by definition of distance.

  9. More Results • Def: A median of a triangle is a line segment that has as its endpoints a vertex of the triangle and the midpoint of the side opposite that vertex. • Theorem 4.2.2: Median Concurrence Theorem (neutral) • The three medians of a triangle are concurrent at a point called the centroid • Cor 4.2.8: Any two medians of a triangle intersect at a point that is 2/3 the distance from any vertex to the midpoint of the opposite side.

  10. More Results • Theorem 4.2.9: Two lines parallel to the same line are parallel to each other. • Proof:Let l, m, n be lines such that l || m, m||n • wts: l||n • Let t be a transversal of n,m,l • Then <1 and <7 are congruent as well as <5 and <11 • Since vertical angles are congruent, <1 ≡<7 ≡ <5 ≡<11 and <1 ≡ <11 • And <1 and <11 are congruent alt. Int. angles, l || n

  11. More Results • Property: Parallel lines are everywhere equidistant • Proof: • Let l || m. Draw PS ┴ m at S and QR┴m at R. Then םPQRS is a parallelogram. • By theorem 4.2.4, PS ≡QR • So parallel lines are equidistant everywhere.

  12. More Results • There are 12 more simple theorems listed on page 134-135 (read)

  13. #8 Prove Theorem 4.2.11 • Each diagonal of a parallelogram partitions the parallelogram into a pair of congruent triangles. • Proof: • Let םABCD be a parallelogram • wts: ΔABC ≡ΔCDA and ΔBAD ≡ΔDBC • By theorem 4.2.4, AB ≡ DC and BC ≡ AD • Draw diagonals BD and AC • By SSS, ΔABC ≡ΔCDA and ΔBAD ≡ΔDBC

  14. #10 Prove Theorem 4.2.13 • If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram • Proof: • Let םABCD be a quadrilateral such that AC bisects BD and BD bisects AC • wts: םABCD is a parallelogram • Know BE ≡ ED, AE ≡ EC, and <1 ≡ <3 and <2 ≡ <4 (vertical angles) • So ΔCED ≡ΔAEB and ΔCEB ≡ΔAED by SAS • Thus, AB ≡ CD and BC ≡ AD • Since opposite sides of םABCD are congruent, by theorem 4.2.4, םABCD is a parallelogram

  15. #15 Prove Theorem 4.2.18 • The median to the hypotenuse of a right triangle is one-half the length of the hypotenuse • Proof: • Let ΔABC be a right triangle with hypotenuse BC • Draw AD such that D is the midpoint of CB. Then AD is a median of ΔABC. • wts: AD =1/2 BC • Draw line AE such that A-D-E and AD ≡ DE • Draw DE and BE • Since AE and BC bisect each other םABEC is a parallelogram (theorem 4.2.12) • So AB ≡ CE and CA ≡ EB (theorem 4.2.4)

  16. cont. • So by theorem 4.2.12, we have ΔABC ≡ΔECB and ΔBAE ≡ΔCEA • which implies <A ≡ <E and <B ≡ <C • This implies m<A=m<E=90 and <B=m<C • Since the sum of the angles of םABEC=360, m<B=m<C=90 • So by SAS, ΔABE ≡ΔBAC • Then CB ≡ EA which implies CB=EA which implies • AD= ½ CB since they bisect each other.

  17. Assign 4, 13, 22, 23

  18. 4.3 Congruence and Area • Postulates 17-20 • 17 (area): Every polygon has a unique area • 18 (area congruence): Congruent triangles (congruent polygons) have the same area • 19 (area addition): Area, as a quantity, is additive in nature • 20 (formula): We can find the area of a rectangle as length*width

  19. Theorem 4.3.1-Parallelograms that share a common base and that have sides opposite this base contained in the same (parallel) line are equal (in area) • Every parallelogram has two altitudes and 2 bases. • Theorem 4.3.2-The area of a parallelogram is the product of the lengths of its base and its height.

  20. Definitions • The height of a triangle is the measure of the perpendicular line segment that is drawn from the base line to the opposite vertex • trapezoid: quadrilateral having two parallel and two non-parallel sides and the height of the trapezoid is the measure of the perpendicular drawn between the parallel sides • rhombus: parallelogram in which all 4 sides are congruent

  21. Theorem 4.3.3-The area of a right triangle is ½ the product of the lengths of its legs • Proof: • Given right ΔABC with m<C=90 • wts: A= ½ ab • Construct l┴BC at B • Construct m┴l from A • Then ABCD is a rectangle. • So, area of ACBD=ab • Also ΔABC ≡ΔBAD, so area ΔABC = area ΔBAD • And area of ACBD= aΔABC + aΔBAD • Then aΔABC= ½ area ACBD = ½ ab

  22. Theorem 4.3.4-The area of a triangle is ½ the product of any base and the corresponding height • Set up proof: • In any triangle, a= ½ bh • Have to show three cases: • If right, each leg is an altitude to the other and leg is the base • If acute, drop the altitude to form two right triangles • If obtuse, drop the altitude down to the extension of one of the sides

  23. Case 1-right triangle • Let ΔABC is right triangle with m<C=90 • wts: a= ½ bh • Since the triangle is right, each leg is a height to the other leg, which is a base. • So for ΔABC, by theorem 4.3.3, aΔABC= ½ ab = ½ bh

  24. Case 2- Acute • Consider acute ΔABC and let CB be the base • Draw AD ┴ CB such that C-D-B • This separates the triangle into 2 right Δ ‘s • Which implies aΔADB= ½ (AD)(DB) and aΔADC= ½ (DC)(AD) • which implies aΔACB= aΔADB + aΔADC = ½ (AD)(DB) + ½ (DC)(AD) = (½AD)(DB+DC) = ½ (AD)(CB) = ½ bh

  25. Theorem 4.3.5- The area of a trapezoid is the product of its height and the arithmetic mean of its bases • Proof: • Draw trapezoid ABCD, such that BC || AD and BC < AD • Since AD > BC, there are two possible cases: • Case 1: AB ┴ AD • Which implies AB ┴ BC • Draw CE ┴ AD such that A-E-D • Then םABCE is a rectangle and ΔCED is right • So, aםABCE = (AB)(AE) and aΔCED= ½ (ED)(CE) • Then aABCD= aםABCE + aΔCED (postulate 19) = (AB)(AE) + ½ (ED)(CE) • Since AB ≡ CE, aABCD = (AB)(AE + ½ ED)= (AB)(BC+AD)/2 • = h(b1 + b2)/2

  26. Definition/notations: n-gon • Assign: #2, 3, (both we did in class), 4, 5, 6, 7, 13, 17, 18. • Turn in 5, 6, 7 Wed October 24 • Also due Monday October 22: One page description of paper

  27. 4.4 Similarity • Similarity: Two polygons ABCD…Z and A’B’C’D’…Z’ are similar if and only if • m<A=m<A’, m<B=m<B’, … , m<Z=m<Z’ • AB = BC = CD = … = ZA A’B’ B’C’ C’D’ Z’A’ • Theorem 4.4.1: Similarity is an equivalence relation

  28. Theorem 4.4.2: If a line parallel to one side of a triangle intersects the other two sides in two different points, then it divides these sides into segments that are proportional. • Read Proof pages 146-148 • Read Cor 4.4.3 • Theorem 4.4.4: If a line l intersects two sides of a triangle in different points so that it cuts off segments that are proportional to the sides, then the line is parallel to the third side. • Read Proof page 148-149

  29. Theorem 4.4.5- AAA Similarity: If for two triangles, all angles are congruent, then the triangles are similar. • Read proof • Finish proof: • Now assume AC ≠ DF, so assume DF > AC • Locate point J between D and F such that DJ=AC • Likewise, locate I on DE s.t. DI = AB. • Since m<A =m<D, ΔABC ≡ ΔDIJ (by SAS) • So then m<IJD = m<C =m<F and thus IJ || EF • Then by Cor 4.4.3: DI/DE = DJ/DF • Since DJ=AC, DI=AB then we have AB/DE = AC/DF

  30. Similarity Theorems • 4.4.6- If for two triangles, an angles of one triangle is congruent to the corresponding angle from the other triangle, and if the corresponding sides that surround this angle are proportional, then the triangles are similar. • 4.4.7- If for two triangles, the lengths of the 3 sides of one triangle are proportional to the lengths of the corresponding 3 sides of the other triangle, then the triangles are similar.

  31. Proof for SAS Similarity • Given ΔABC and ΔXYZ • Let <A ≡ <X and AB/XY = AC/XZ • wts: ΔABC ~ ΔXYZ • Assume m<C > m<X • Then by angle construction postulate, there exists <YZW such that <YZW ≡ <C and Y-W-X • By AAA, ΔABC ≡ΔWYZ which implies • AB/WY = AC/WZ which is a contradiction unless W=X

  32. Proof for SSS Similarity • Given ΔABC and ΔDEF such that AB/DE = BC/EF= AC/DF • wts: ΔABC ~ ΔDEF • Find E’ on AB such that AE’=DE and find F’ on AC such that AF’=DF • Then AB/AE’ =AC/AF’ by substitution • So ΔABC ~ ΔAE’F’ since <A ≡ <A (SAS) • So E’F’/BC=AE’/AB • So E’F’ = (BC)(AE’)/AB = (BC)(DE)/AB • And EF = (BC)(DE)/AB • So EF= E’F’ • So ΔAE’F’ ≡ΔDEF by SAS • Hence <A = <D and • ΔABC ~ ΔDEF by SAS

  33. Similarity is also use to prove the Pythagorean Theorem (4.4.8) • Do number 11 • Assign 1, 3, 10, 12, 13, 14 • Turn in 1, 10, 14 Monday October 29

  34. 4.5-Euclidean Results About Circles • Definition: A circle is the set of all points equidistant from a given point • The given point is called the center and the common distance is the radius • Theorem 4.5.1- In the Euclidean plane, three distinct, non-collinear points determine a unique circle • Theorem 4.5.2-Every triangle can be circumscribed, and the center of the circumscribing circle is the concurrence point of the perpendicular bisectors of two of the sides of the triangle.

  35. Def: A chord of a circle is a line segment joining two points on the circle • Def: A diameter of a circle is a chord that contains the center of the circle • Theorem 4.5.3: If AB is a diameter of a circle and if CD is another chord of the same circle that is not the diameter then AB > CD • Read Proof Page 160

  36. Theorem 4.5.4: If a diameter of a circle is perpendicular to a chord of the circle, then the diameter bisects the chord. • Theorem 4.5.5: If a diameter of a circle bisects a chord of the circle (which is not a diameter), then the diameter is perpendicular to the chord. • Theorem 4.5.6: The perpendicular bisector of a chord of a circle contains a diameter of the circle.

  37. Proof Theorem 4.5.5 • Let AB be a diameter of C(O, OA) • Let CD be a chord, not the diameter on circle • Let AB bisect CD at point P • wts: AB ┴CD • Draw OC and OD. Since C and D are on the circle, then OC ≡ OD • We know OP ≡ OP and we are given CP≡PD • So by SSS ΔCPO ≡ΔDPO • which implies <OPC ≡ <OPD and since they form a linear pair both must be 90. • So AB ┴CD

  38. Other Definitions • A secant of a circle is a line that contains exactly 2 points of the circle • A tangent of a circle is a line that contains exactly one point of the circle • Theorem 4.5.7: If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency. • Read proof page 162 • Any angle whose vertex is the center of a circle is called a central angle for the circle

  39. #7 • Given C(O,OA) and P in the exterior of circle • Draw tangents PT and PS such that T and S lie on circle C(O,OA) • wts: PT ≡ PS • Draw OS and OT • By theorem 4.5.7, <PSO and <PTO are right angles. • Draw OP • Since OP ≡ OP and OS ≡ OT, then by hypotenuse-leg congruence condition ΔOPT ≡ΔOPS • So PT ≡ PS

  40. #8 • Suppose t is tangent to C(O,OP) at P • Also suppose we have secant l, which is parallel to t, intersects C(O,OP) at the two points A and B • wts: AP ≡ BP • By theorem 4.5.7 OP ┴ t which implies diameter DP ┴ t • Let the intersection of DP and AB be M • By the converse of the Alt. Int. Angle theorem, <BMP ≡ <CPM, which implies m<BMP=90 • So then m<PMA=90 since it is supplementary • And by Theorem 4.5.4, BM ≡ MA • So by SAS ΔBMP ≡ΔAMP • Therefore BP ≡ AP

  41. #9 • Given C(O, OA) • Let DA be a diameter of the circle • Draw line t perpendicular to AD at A • Let C be any other point on t • Draw OC • Since the perp segment from a pt to a line is the shortest segment from the pt to a line, OC > OA • Therefore, C lies in the exterior of the circle • Hence, t is tangent to C(O,OA).

  42. #11 • Given C(O,OA), such that OX┴AB, OY┴CD, and OX=OY • Draw OD, OC, OA, and OB • we know that OD≡OC≡OA≡OB since they are all radii • Then ΔOCY≡ΔODY≡ΔOAX≡ΔOBX by the hypotenuse leg condition • Therefore, YC≡YD≡XA≡XB or YC=YD=XA=XB • so YC+YD=XA+XB • which implies CD=AB, so CD≡AB

  43. Definitions • Draw example of semicircle, minor arc and major arc • Draw example of inscribed angles and intercepted arcs and measure of angles. • Theorem 4.5.8:If two chords of a circle are congruent, then their corresponding minor arcs have the same measure (see pic page 165) • Converse: If two minor arcs are congruent, then so are the corresponding chords

  44. #17 proof • Let arcAB ≡ arcCD • wts: AB ≡ CD • We know by definition that <AOB ≡ <COD • Draw OA, OB, OC, and OD • Since they are radii, OA≡OB≡OC≡OD • Draw AB and CD • Then by SAS ΔCOD ≡ΔAOB • Therefore, AB≡CD

  45. Theorem 4.5.11: The measure of an angle inscribed in an arc is ½ the measure of its intercepted arc. • Proof: • Draw OA, then m<AOC=m arcAC • m<1=m arcAC • m<1=m<2+m<3 • So, m<1=2m<3 • Then, m<3= ½ m<1= ½ m arcAC

  46. Theorem 4.5.14: If two chords intersect in the interior of a circle to determine an angle, the measure of the angle is the average of the measure of the arcs intercepted by the angle and its vertical angles. • Theorem 4.5.15: If two secants intersect at a point in the exterior of a circle, the measure of the angle at the point of intersection is ½ the positive difference of the two intercepted arcs.

  47. Suppose we are given line AB is tangent to C(O,OA) at point A and AC is a chord such that m arcAPC=xº and O is an interior point of <BAC. Then what is m<BAC? • Well m<BAC=90+ ½ m arcCD • = ½ m arcAPD + ½ m arcCD • = ½ (m arcAPD + arcCD) • = ½ m arcAPC

  48. Theorem 4.5.16: If line AB is tangent to C(O,OA) at point A and if AC is a chord such that m arcAPC=x, then m<BAC= ½ xº • Proof: (#25) when AC contains the center: • wts: xº= ½ m arcAPC • It has already been shown that m<BAP = ½ m arcAP • m<PAC= ½ m arcPC • xº= m <BAP +m<PAC = ½ m arcAP + ½ m arc PC • = ½ (m arcAP + arcPC)= ½ m arcAPC

  49. Homework: 10, 12, 14, 22, 24, 26, 27, 30, 31 • Look at #24 if one of the secants is the center of the circle: • Given line AC a secant of C(O,OC) such that A-O-C and AD a secant of C(O,OC) such that AC and AD intersect on the exterior of circle at A • Draw EC. Then m<BCE= ½ arcBE and m<CED= ½ arcCD. • We know m<CAD+m<ACE=m<CED • So m<CAD=m<CED- m<ACE • By substitution, m<CAD= ½ (arcCD – arcBE)

  50. 4.6 Euclidean Results of Triangles • Already have: • The three medians of a triangle are concurrent (at a point called the centroid) • The three perpendicular bisectors of a triangle are concurrent (at a point called the circumcenter) • Shortest distance from a point to a line is the perpendicular dropped from the point to the line. • From this we have: A point is on the bisector of an angle if and only if it is equidistant from the sides of the angle.

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