MST Topics in History of Mathematics

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1 MST Topics in History of Mathematics Euclid s Elements and the Works of rchimedes Paul Yiu Department of Mathematics Florida tlantic University Summer 2014 June 30

2 2.6 ngle properties ngle properties ngle between lines Euclid (I.13). If a straight line stands on a straight line, it makes either two right angles or angles equal to two right angles. Euclid (I.14). If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines are in a straight line with one another. Euclid (I.15). If two straight lines cut one another, they make the vertical angles equal to one another. Porism. From this it is manifest that if two straight lines cut one another, they will make the angles at the point of section equal to four right angles.

3 12 Euclid s Elements, ook I (Propositions) ngles of a triangle Euclid (I.16). In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles. F E D G Given: Triangle with the side extended to D. To prove: The exterior angle D is greater than either of the interior and opposite angles,. onstruction: isect at E; [I.10] Join E and extend to F such that EF = E, [I.3] Join F and extend to G. [Post. 1,2] Proof : In triangles E and EF, E = E E = FE E = EF E EF E = EF. ut ED > EF D(> ED) > E =. Similarly, G > angle. ut D = G D >. construction construction [I.15] [I.4] [.N.5] [I.15]

4 2.6 ngle properties 13 Euclid (I.17). In any triangle two angles taken together in any manner are less than two right angles. The famous angle sum theorem of a triangle depends on the parallel postulate. Euclid (I.32). In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.

5 14 Euclid s Elements, ook I (Propositions) 2.7 Inequalities Euclid (I.18). In any triangle the greater side subtends the greater angle. D Given: Triangle with greater than. To prove: >. onstruction: Point D on such that D = ; Join D. Proof : ngle D is an exterior angle of triangle D, D > D (interior opposite angle). = D D = D D > ; >. [I.16] construction [I.5] [I.3]

6 2.7 Inequalities 15 Euclid (I.19). In any triangle the greater angle is subtended by the greater side. Given: Triangle with >. To prove: >. Proof by contradiction: (1) Suppose =. Then =. [I.5] This contradicts the assumption. (2) Suppose <. onstruct a point D on such that D =. [I.3] Triangle D is isosceles. D = D. [I.5] ut < D [I.16] D = D (Triangle D is isosceles). D <. <, a contradiction. (3) Therefore, >. Q.E.D.

7 16 Euclid s Elements, ook I (Propositions) Euclid (I.20). [Triangle inequality] In any triangle two sides taken together are greater than the remaining one. Given: Triangle. To prove: + >. onstruction: Extend to D such that D =. Join D. Proof : In triangle D, D > D = D = D. D >. ut D = + D = +. + >. Q.E.D.

8 2.7 Inequalities 17 Euclid (I.21). If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle. Given: Triangle and two line D, D with D inside the triangle. To prove: (i) D + D < +, and (ii) D >. onstruction: Extend D to intersect at E. Proof : (i) D + D < D +(ED + E) triangle inequality = (D + ED)+E = E + E triangle inequality < ( + E)+E = +(E + E) = +. triangle inequality (ii) D > E > [I.16]

9 18 Euclid s Elements, ook I (Propositions) 2.8 Steiner-Lehmus theorem Theorem 2.1. triangle is isosceles if and only if it has two equal internal angle bisectors. Lemma 2.2. Let triangles and XY Z be equiangular. If > XY, then > XZ and > Y Z. Z Given: Equiangular triangles and XY Z with > XY. To prove: > XZ and > Y Z. onstruction: Let be a point on such that = XY. onstruct the parallel through to, intersecting the line at. Proof : must be between and, for otherwise, and are on opposite sides of, and the lines, intersect. This contradicts their parallelism. Now, triangles and XY Z are congruent. [S] Therefore, > = XZ. X Y

10 2.8 Steiner-Lehmus theorem 19 Given: Triangle with <and bisectors E, F. To prove: E > F. onstruction: Point G on such that GF = E. Join G and let it intersect F at H. G F H E Proof : The triangles GH and GF are equiangular with G > G. Therefore, H > F. (Lemma above) Since E > H, E > F.

11 1 onstruction Problem 1. Given a segment, construct an isosceles triangle with = = 2.

12 2 Solution to onstruction Problem 1. Given a segment, construct an isosceles triangle with = = 2.

13 onstruction Problem 2. Given the center of a circle, find two opposite points on the circumference by using a compass only. 3

14 4 Solution to onstruction Problem 2. Given the center of a circle, find two opposite points on the circumference by using a compass only. O

15 onstruction Problem 3. onstruct a square with two given points as opposite vertices. 5

16 6 Solution to onstruction Problem 3. onstruct a square with two given points as opposite vertices. O

17 7 onstruction Problem 4. Given triangle, construct a line DE parallel to the base to intersect and at D and E respectively such that DE = D + E. D E

18 8 Solution to onstruction Problem 4. Given triangle, construct a line DE parallel to the base to intersect and at D and E respectively such that DE = D + E. D E

19 9 onstruction Problem 5. Given two lines and a point not on any of them, construct a line through to intersect the given lines at and such that =.

20 10 Solution to onstruction Problem 5. Given two lines and a point not on any of them, construct a line through to intersect the given lines at and such that =.

21 11 onstruction Problem 6. onstruct triangle,givena,, and b + c. a c b

22 12 onstruction Problem 7. onstruct triangle, givena,, and b c. b c a

23 13 onstruction Problem 8. onstruct triangle,given,, and the perimeter a + b + c. c a b

24 14 onstruction Problem 9. Find the point in the base of a triangle from which lines construct parallel to the sides to meet them are equal.

25 15 onstruction Problem 10. Trisect a given segment. Given a segment, and a point not on the line containing, (i) extend to D so that = D, (ii) take the midpoint E of, (iii) construct the line DE to intersect at F. Then =3 F. D E F

26 16 Solution to onstruction Problem 10. Trisect a given segment. Given a segment, and a point not on the line containing, (i) extend to D so that = D, (ii) take the midpoint E of, (iii) construct the line DE to intersect at F. Then =3 F. D G E onstruction: onstruct the parallel of through to intersect the line DF at G. Proof : (1) In triangles F E and GE, FE = GE, (alternate angles between parallel lines) E = E, (midpoint) EF = EG. (I.15) F E GE. (S) F = G. (2) In triangle DF, G//F and is the midpoint of D. G is the midpoint of DF, (Intercept theorem) and F =2G. (Midpoint theorem) From (1) and (2) F =2F = =3F. F

27 hapter 3 Euclid s Elements, ook II 3.1 II.1-8 Definitions. (II.1). ny rectangular parallelogram is said to be contained by the two straight lines containing the right angle. (II.2). nd in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a gnomon. Euclid (II.1). If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments.

28 2 Euclid s Elements, ook II Euclid (II.2). If a straight line be cut at random, the rectangles contained by the whole and both of the segments is equal to the square on the whole. (II.3). If a straight line be cut at random, the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment.

29 3.1 II Euclid (II.4). If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments. (a + b) 2 = a 2 + b 2 +2ab. L D H G K K L H M D F E E G F (II.5). If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half.

30 4 Euclid s Elements, ook II Euclid (II.6). If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.

31 3.1 II Euclid (II.7). If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment and the square on the remaining segment. Euclid (II.8). If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line.

32 6 Euclid s Elements, ook II 3.2 II.9-14 Euclid (II.9). If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and the square on the straight line between the points of section. E E F G F M D D G Euclid (II.10). If a straight line be bisected, and a straight line be added to it in a straight line, the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and the square described on the straight line made up of the half and the added straight line as on one straight line.

33 3.2 II Euclid (II.11). To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment. F G H E Given: segment. To construct: point H on such that the square on H is equal to the rectangle contained by and H. onstruction: onstruct a square D, and let E be the midpoint of. Extend to F such that EF = E. onstruct a square F GH with H on. To prove: The square on H is equal to the rectangle contained by and H. Proof : Since E is the midpoint of, and F GH is a square, the rectangle contained by F and FG, together with the square on E, is the square on EF. (II.6) In the right triangle E, the square on, together with the square on E, is the square on E. (I.47) Since EF = E (by construction), the rectangle contained by F and FGis equal to the square on. Therefore, the square F GH is equal to the rectangle DKH, which is equal to the rectangle contained by and H. K D

34 8 Euclid s Elements, ook II Euclid (II.11). To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment. F G H E Given: segment. To construct: point H on such that the square on H is equal to the rectangle contained by and H. onstruction: onstruct a square D, and let E be the midpoint of. Extend to F such that EF = E. onstruct a square F GH with H on. To prove: The square on H is equal to the rectangle contained by and H. Proof without words: K D F G F G H H E E K D K D F G F G H H E E K D K D

35 8 Euclid s Elements, ook II Euclid (II.12). In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle. D D Euclid (II.13). In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle. Proof. (1) (2)

36 3.2 II orollary (pollonius theorem) If D is the midpoint of the side of triangle, =2(D 2 + D 2 ).

37 10 Euclid s Elements, ook II 3.3 onstruction of a square equal to a rectangle Euclid (II.14). To construct a square equal to a given rectilinear figure. G F E H D rectilineal figure can be made equal to a rectangle (I.45). Proof. G F E H D G F E H D

Chapter 1. Euclid s Elements, Book I (constructions)

Chapter 1. Euclid s Elements, Book I (constructions) hapter 1 uclid s lements, ook I (constructions) 102 uclid s lements, ook I (constructions) 1.1 The use of ruler and compass uclid s lements can be read as a book on how to construct certain geometric figures