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 efficiently and accurately using ruler and compass, and ascertaining the validity. The first three postulates before ook I are on the basic use of the ruler and the compass. Postulate 1. To draw a straight line from any point to any point. With a ruler (straightedge) one connects two given points and to form the line (segment), and there is only one such line. This uniqueness is assumed, for example, in the proof of I.4. Postulate 2. To produce a finite straight line continuously in a straight line. Given two points and, with the use of a ruler one can construct a point so that the line (segment) contains the point. The first two postulates can be combined into a single one: through two distinct points there is a unique straight line. Postulate 3. To describe a circle with any center and distance. This distance is given by a finite line (segment) from the center to another point. With the use of a collapsible compass, one constructs a circle with given center to pass through. We denote this circle by (, ). uclid I.2 shows how to construct a circle with a given center and radius equal to a given line (segment). efinition (I.20). Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal. 1 uclid (I.1). On a given finite straight line to construct an equilateral triangle. Given: Points and. To construct: quilateral triangle. onstruction: onstruct the circles (, ) and (,) to intersect at a point. is an equilateral triangle. 1 uclid seems to take isosceles and scalene in the exclusive sense. ut it is more convenient to take these in the inclusive sense. n isosceles triangle is one with two equal sides, so that an equilateral triangle is also isosceles.
1.1 The use of ruler and compass 103 The second proposition is on the use of a collapsible compass to transfer a segment to a given endpoint. uclid (I.2). To place a straight line equal to a given straight line with one end at a given point. L G Given: Point and line. To construct: Line L equal to. onstruction: (1) n equilateral triangle, [I.1] (2) the circle (,). (3) xtend to intersect the circle at G. (4) onstruct the circle (, G) and (5) extend to intersect this circle at L. L = L = G = G =. Therefore the circle (, ) can be constructed using a collapsible compass. uclid (I.3). Given two unequal straight lines, to cut off from the greater a straight line equal to the less. uclid (I.9). To bisect a given rectilineal angle. F Given: ngle. To construct: Line F bisecting angle. onstruction: (1) hoose an arbitrary point on. (2) onstruct on such that =. [I.3] (3) onstruct an equilateral triangle F on (so that F and are on opposite sides of ). [I.1] The line F is the bisector of the angle.
104 uclid s lements, ook I (constructions) uclid (I.10). To bisect a given finite straight line. Given: Line segment. To construct: The midpoint of. onstruction: (1) n equilateral triangle, [I.1] (2) the bisector of angle [I.9] (3) to meet at. =. 1.2 Perpendicular lines efinition (I.10). When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. Postulate 4. That all right angles are equal to each other. uclid (I.11). To draw a straight line at right angles to a given straight line from a given point on it. F Given: straight line and a point on it. To construct: line F perpendicular to. onstruction: (1) Take an arbitrary point on, and construct on such that =. (2) onstruct an equilateral triangle F. [I.1] The line F is perpendicular to. Proof : In triangles F and F, =, by construction F = F, F = F sides of equilateral triangle F F SSS F = F F.
1.3 Isosceles triangles 105 uclid (I.12). To draw a straight line perpendicular to a given infinite straight line from a given point not on it. G H Given: straight line and a point not on the line. To construct: line H perpendicular to. onstruction: (1) Take a point on the other side of the line and construct the circle (, ) to meet the line at and G. isect the segment G at H [I.10] H is perpendicular to. Proof : Triangles H and GH are congruent. [SSS] H = GH. H. 1.3 Isosceles triangles uclid (I.5). In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another. F G Given: Triangle with =. To prove: =. onstruction: xtend and to and respectively. hoose an arbitrary point F on, and construct G on such that G = F. [I.3]
106 uclid s lements, ook I (constructions) Proof : In triangles F and G, =, F = G, F = + F = + G = G. G F, F = G, and F = F = G = G. In triangles F and G, =, F = G, F = G, F G F = G =. isosceles triangle SS proved above SS Q... 2 This proposition is often called the bridge of asses (pons asinorum). ccording to Smith 3, [i]t is usually stated that it came from the fact that fools could not cross this bridge, and it is a fact that in the Middle ges this was often the limit of the student s progress in geometry. It has however been suggested that the name came from uclid s figure, which resembles the simplest type of a wooden truss bridge. ccording to Proclus, Pappus (3rd century..) proved the proposition in a very simple way by considering an isosceles triangle as two triangles, and. From =, =, =, it follows from I.4 that, and so =. Here is another apparently easy proof of I.5: let the bisector of angle meet at. Then by I.4 again. From this we conclude that =, equivalently, =. The existence of an angle bisector is justified only in I.9. uclid is a true constructivist. He does not consider squares before I.46. uclid (I.6). If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another. Given: Triangle with =. To prove: =. Proof by contradiction: 4 (1) Suppose is greater than. 2 quod erat demonstrandum, which was to be demonstrated. 3.. Smith, The Teaching of Geometry, 1911, p.174. 4 reductio ad absurdum.
1.4 Tests for congruence of triangles I.4,8,26 107 hoose a point on such that =. [I.3] In triangles and, =, =, =.. SS This is contradiction since triangle is part of triangle. (2) Suppose is less than. similar reasoning also leads to a contradiction. (3) Therefore, =. Q... 1.4 Tests for congruence of triangles I.4,8,26 uclid did not use the term congruence of triangles. When he says two triangles are equal, he means they are equal in area. uclid (I.4). [SS] If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles equal the remaining angles respectively, namely those which the equal sides subtend. It is convenient to identify congruent triangles with corresponding vertices and sides. Thus, two triangles and XY Z are congruent if the corresponding sides are equal, namely, = YZ, = ZX, XY =, and the corresponding angles are equal, namely, = YXZ, = ZY X, = XZY. uclid I.4 is the first of four valid tests for congruence of triangles. We refer to it as the SS test. ook I contains two more tests. uclid (I.8). [SSS] If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will have the angles equal which are contained by the equal straight lines. uclid (I.26). [S, S] If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.
108 uclid s lements, ook I (constructions) The following proposition, though not in uclid s lements, completes the list of congruence tests. Proposition (RHS). Two right angled triangles with equal hypotenuses and one pair of equal sides are congruent. This is, of course, is a corollary of I.48. It can, however, be established without invoking this. Let and be two triangles in which the angles and are right angles, = and =. On the side of opposite to, construct a triangle congruent to (with = and = ). Since the angles and are both right angles,,, are a line. Since =, by I.5, =. Therefore, triangles and are congruent (I.4); so are and. 1.5 Two constructions uclid (I.22). Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. uclid (I.23). [opying an angle along a line] On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectlineal angle. Given: line and an angle. To construct: F equal to angle. onstruction: hoose arbitrary points, on the sides of the angle. onstruct triangle F G such that = F, = G and = FG. [I.22] 1.6 Parallel lines efinition (I.23). Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. Postulate 5. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. This famous fifth postulate has a long history, eventually leading to the discovery of noneuclidean geometry in the 19th century. Here, we simply remark that it provides an explicit criterion for the intersection of finite lines upon extension. Heath [I.201] pointed out that while Postulate 4 can be proved using other axioms, [i]t was essential from uclid s point of view that it should come before Post. 5, since the condition in the latter that a certain pair of angles are together less than
1.7 Parallelograms 109 two right angles would be useless unless it were first made clear that right angles are angles of determinate and invariable magnitude. uclid (I.27). If a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another. (I.28). If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another. (I.29). straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the sum of the interior angles on the same side equal to two right angles. (I.30). Straight lines parallel to the same straight line are also parallel to one another. uclid (I.31). Through a given point to draw a straight line parallel to a given straight line. Given: point and a line not containing it. To construct: line through parallel to. onstruction: (1) Take a point on. (2) opy angle into angle with and on opposite sides of. The line is parallel to. 1.7 Parallelograms uclid (I.33). The straight lines joining equal and parallel straight lines [at the extremities which are] in the same directions are themselves equal and parallel. uclid (I.34). In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.
110 uclid s lements, ook I (constructions) uclid (I.35). Parallelograms which are on the same base and in the same parallels are equal to one another. F uclid (I.36). Parallelograms which are on equal bases and in the same parallels are equal to one another. (I.37). Triangles which are on the same base and in the same parallels are equal to one another. (I.38). Triangles which are on equal bases and in the same parallels are equal to one another. (I.39). qual triangles which are on the same base and on the same side are also in the same parallels. (I.40). qual triangles which are on equal bases and on the same side are also in the same parallels. (I.41). If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle. uclid (I.42). To construct, in a given rectilineal angle, a parallelogram equal to a given triangle. F G uclid (I.43). In any parallelogram the complements of the parallelograms about the diameter are equal to one another.
1.8 The Pythagorean theorem 111 H K F G uclid (I.44). To a given straight line to apply, 5 in a given rectilineal angle, a parallelogram equal to a given triangle. F K G M H L uclid (I.45). To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure. F G L K H M 1.8 The Pythagorean theorem efinition (I.22). Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. nd let quadrilaterals other than these be called trapezia. uclid (I.46). To describe a square on a given straight line. Given: line. To construct: square. onstruction: (1) onstruct the circle (, ). (2) onstruct the perpendicular to at to meet the circle at. (3) onstruct the circles (,) and (, ) to meet at. is a square. uclid (I.47). In right angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.
112 uclid s lements, ook I (constructions) K K G G H H F F L L ook I ends with the converse of I.47. uclid (I.48). If in a triangle the square on one of the sides be equal to the squares on the remaining sides of the triangle, the angle contained by the remaining sides of the triangle is right. Given: Triangle with 2 + 2 = 2. To prove: =1rt.. onstruction: onstruct triangle with =1rt. and =, =. Proof. 2 = 2 + 2 = 2 + 2 = 2. =. In triangles and, =, =, =, [SSS] and = =1rt.. Q... ecause of its usefulness, we reformulate the key step of uclid s proof of I.47 as follows. 5 The term application is explained in Proclus commentary: application starts with one side given and constructs the area along it, neither falling short of the length of the line nor exceeding it, but using it as one of the sides enclosing the area.
1.8 The Pythagorean theorem 113 Proposition. Let the perpendicular from the right angle vertex of a right triangle intersect the hypotenuse at X. Then 2 = X, 2 = X. X uclid (II.14). To construct a square equal to a given rectilineal figure. H G F Proof. (1) rectilineal figure can be made equal to a rectangle [I.45]. (2) onversion of a rectangle into a square: H H G F G F
114 uclid s lements, ook I (constructions) 1.9 ppendix: Tests of congruence of triangles onstruction of a triangle with three given elements triangle has six elements: three sides and three angles. onsider the construction of a triangle given three of its six elements. The triangle is unique (up to size and shape) if the given data are in one of the following patterns. (1) SSS Given three lengths a, b, c, we construct a segment with length a, and the two circles (c) and (b). These two circles intersect at two points if (and only if) b + c>a[triangle inequality]. There are two possible positions of. The resulting two triangles are congruent. (2) SS Given b, c, and angle <180, the existence and uniqueness of triangle is clear. (3) S or S These two patterns are equivalent since knowing two of the angles of a triangle, we easily determine the third (their sum being 180 ). Given, and a, there is clearly a unique triangle provided + <180. (4) RHS Given a, c and =90. ongruence tests These data patterns also provide the valid tests of congruence of triangles. Two triangles and XY Z are congruent if their corresponding elements are equal. Two triangles are congruent if they have three pairs of equal elements in one the five patterns above. The five valid tests of congruence of triangles are as follows. (1) SSS: XY Z if = XY, = YZ, = ZX. X Z (2) SS: XY Z if Y = XY, = XY Z, = YZ. X Z Y
1.9 ppendix: Tests of congruence of triangles 115 X Z Y (3) S: XY Z if = YXZ, = XY, = XY Z. (4) S. We have noted that this is the same as S: XY Z if = YXZ, = XY Z, = YZ,. X Z Y (5) RHS. The SS is not a valid test of congruence. Here is an example. The two triangles and XY Z are not congruent even though = YXZ, = XY, = YZ. Y X However, if the equal angles are right angles, then the third pair of sides are equal: 2 = 2 2 = YZ 2 XY 2 = XZ 2, and = XZ. The two triangles are congruent by the SSS test. Without repeating these details, we shall simply refer to this as the RHS test. XY Z if = YXZ =90, = YZ, = XY. Z Y Z X
116 uclid s lements, ook I (constructions)
hapter 2 uclid s lements, ook I (Propositions) 2.1 efinitions efinitions. (I.1). point is that which has no part. (I.2). line is breadthless length. (I.3). The extremities of a line are points. (I.4).Ȧ straight line is a line which lies evenly with the points on itself. efinitions. (I.5). surface is that which has length and breadth only. (I.6). The extremities of a surface are lines. (I.7). plane surface is a surface which lies evenly with the straight lines on itself. efinitions. (I.8). plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line. (I.9). nd when the lines containing the angle are straight, the angle is called rectilineal. efinition (I.10). 1 When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. efinitions. (I.11). n obtuse angle is an angle greater than a right angle. (I.12). n acute angle is an angle less than a right angle. 1 See 1.2.
118 uclid s lements, ook I (Propositions) efinitions. (I.13). boundary is that which is an extremity of anything. (I.14). figure is that which is contained by any boundary or boundaries. efinitions. (I.15). circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure equal one another. (I.16). nd the point is called the center of the circle. (I.17). diameter of the circle is any straight line drawn through the center and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle. (I.18). semicircle is the figure contained by the diameter and the circumference cut off by it. nd the center of the semicircle is the same as that of the circle. efinition (I.19). Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines. efinitions. (I.20). Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal. (I.21). Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acute-angled triangle that which has its three angles acute. efinition (I.22). Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. nd let quadrilaterals other than these be called trapezia. efinition (I.23). 2 Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. 2.2 The common notions Preceding the five common notions are five postulates. See 1.1, 1.2, 1.6. 2 See 1.6.
2.3 The propositions 119 ommon Notions. 1. Things which are equal to the same thing are also equal to one another. 2. If equals be added to equals, the wholes are equal. 3. If equals be subtracted from equals, the remainders are equal. 4. Things which coincide with one another are equal to one another. 5. The whole is greater than the part. 3 2.3 The propositions I.1-3: the use of ruler and compass. I.4: SS. I.5,6: isosceles triangles. I.7,8: SSS. I.9,10: construction of bisectors of angle and line. I.9-12: construction of perpendicular lines. I.13-15: angle properties. I.16,17: angles of a triangle. I.18,19: comparison of sides in terms of angles, and vice versa. I.20,21: triangle inequality. I.22: construction of triangle given three sides. I.23: copying an angle. I.24,25: comparison of triangles with two pairs of equal sides. I.26: S and S. I.27-30: parallel lines. I.31: construction of parallel line. I.32: external angle of a triangle. I.33,34: parallelograms. 3 For an example of the use of this common notion, see the proof of uclid I.6.
120 uclid s lements, ook I (Propositions) I.35-41: parallelograms and triangles between parallel lines and on common (or equal) bases. I.42-45: constructions: I.46: construction of a square. I.47,48: the Pythagorean theorem and its converse. 2.4 uclid I.4 uclid (I.4). If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. Proof. Let, F be two triangles having... = and = F, and = F. I say that... = F, the triangle equal to triangle F, and = F, = F. For if the triangle be applied to the triangle F, and if the point be placed on the point, and the straight line on, then the point will also coincide with, because =. gain, coinciding with, the straight line will also coincide with F, because = F. Hence the point will also coincide with the point F, because = F. ut also coincide with ; hence the base will coincide with the base F, (For if, when coincides with and with F, the base does not coincide with the base F, two straight lines will enclose a space; which is impossible. Therefore, the base will coincide with F) and will be equal to it. [.N.4] Thus, the whole triangle will coincide with the whole triangle F, and will be equal to it. nd the remaining angles will also coincide with the remaining angles and will be equal to them, = F, and = F. Therefore etc.
2.5 ongruence of triangles: uclid I.7,8;24,25 121 2.5 ongruence of triangles: uclid I.7,8;24,25 uclid (I.7). Given two straight lines constructed on a straight line [from its extremities] and meeting in a point, there cannot be constructed on the same straight line [from its extremities], and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each equal to that which has the same extremity with it. (I.8). If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will have the angles equal which are contained by the equal straight lines. uclid (I.24). If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base. (I.25). If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other. 2.6 ngle properties 2.6.1 ngle between lines uclid (I.13). If a straight line stands on a straight line, it makes either two right angles or angles equal to two right angles. uclid (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. uclid. (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.
122 uclid s lements, ook I (Propositions) 2.6.2 ngles of a triangle uclid (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 G Given: Triangle with the side extended to. To prove: The exterior angle is greater than either of the interior and opposite angles,. onstruction: isect at ; [I.10] Join and extend to F such that F =, [I.3] Join F and extend to G. [Post. 1,2] Proof : In triangles and F, = = F = F F = F. ut > F (> ) > =. Similarly, G > angle. ut = G >. construction construction [I.15] [I.4] [.N.5] [I.15] uclid (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. uclid (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.
2.7 Inequalities 123 2.7 Inequalities uclid (I.18). In any triangle the greater side subtends the greater angle. Given: Triangle with greater than. To prove: >. onstruction: Point on such that = ; Join. Proof : ngle is an exterior angle of triangle, > (interior opposite angle). = = > ; >. [I.16] construction [I.5] [I.3] uclid (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 on such that =. [I.3] Triangle is isosceles. =. [I.5] ut < [I.16] = (Triangle is isosceles). <. <, a contradiction. (3) Therefore, >. Q... uclid (I.20). [Triangle inequality] In any triangle two sides taken together are greater than the remaining one. Given: Triangle. To prove: + >. onstruction: xtend to such that =. Join.
124 uclid s lements, ook I (Propositions) Proof : In triangle, > = =. >. ut = + = +. + >. Q... uclid (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, with inside the triangle. To prove: (i) + < +, and (ii) >. onstruction: xtend to intersect at. Proof : (i) + < +( + ) triangle inequality = ( + )+ = + triangle inequality < ( + )+ = +( + ) = +. triangle inequality (ii) > > [I.16]
hapter 3 uclid s lements, ook II 3.1 II.1-8 efinitions. (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. uclid (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. uclid (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.
126 uclid s lements, ook II uclid (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. L H G K K L H M F 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. uclid (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. uclid (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.
3.2 II.9-14 127 uclid (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. 3.2 II.9-14 uclid (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. F G F M G uclid (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. uclid (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 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, and let be the midpoint of. xtend to F such that F =. K
128 uclid s lements, ook II 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 is the midpoint of, and F GH is a square, the rectangle contained by F and FG, together with the square on, is the square on F. (II.6) In the right triangle, the square on, together with the square on, is the square on. (I.47) Since F = (by construction), the rectangle contained by F and FGis equal to the square on. Therefore, the square F GH is equal to the rectangle KH, which is equal to the rectangle contained by and H. uclid (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. uclid (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)
3.3 onstruction of a square equal to a rectangle 129 orollary (pollonius theorem) If is the midpoint of the side of triangle, 2 + 2 =2( 2 + 2 ). 3.3 onstruction of a square equal to a rectangle uclid (II.14). To construct a square equal to a given rectilinear figure. H G F Proof. rectilineal figure can be made equal to a rectangle (I.45). H H G F G F
130 uclid s lements, ook II 3.4 onstruction exercises for ooks I and II 1. Find a point in a given straight line such that its distances from two given points may be equal. 2. Through two given points on opposite sides of a given straight line, draw two straight lines which intersect on the given line, and include an angle bisected by that given line. 3. onstruct the triangle, given the base, one of the angles at the base, and the sum of the other two sides. 4. is an isosceles triangle with =. Find points on and on such that,, and are all equal. 5. onstruct a right triangle given its hypotenuse and the sum of the other two sides. 6. Given triangle, construct a triangle of equal area, having its vertex at a given point in, and its base in the same straight line as. 7. isect a given triangle by a straight line drawn through a given point in a side. 8. ivide a given straight line into two parts so that the square on one is double the square on the other. 9. ivide a given straight line into two parts so that the sum of their squares is equal to a given square. 10. ivide a given straight line into two parts so that the rectangle contained by them is equal to the square described on a straight line which is less than half of the straight line to be divided.
hapter 4 Modern reorganization of uclid s ook I Theorem 1 (uclid I.13) If a straight line stands on another straight line, the sum of the adjacent angles so formed is equal to two right angles. a + b =2rt. s. bbreviation: djacent angles on a straight line. Theorem 2 (uclid I.14) If the sum of two adjacent angles is equal to two right angles, the exterior arms of the angles are in the same straight line. bbreviation: adj. angles supp. a b Theorem 3 (uclid I.15) If two straight lines intersect, the vertically opposite angles are equal. a = b and p = q. bbreviation: vertically opposite angles. p a b q
132 Modern reorganization of uclid s ook I Theorem 4 (uclid I.4) Two triangles are congruent if they have two pairs of equal sides, and the angles between the pairs of sides are equal: XY Z if X Y Z = XY, = XY Z, = YZ. bbreviation: SS Theorem 5 (uclid I.27,28) Two lines are parallel if a transversal makes (i) a pair of alternate angles equal, or (ii) a pair of corresponding angles equal, or (iii) a pair of interior angles on one side of the transversal supplementary. b a d c bbreviation: (i) alt. angles equal; (ii) corr. angles equal; (iii) int. angles supp. Theorem 6 (uclid I.29) If a transversal cuts two parallel lines, then (i) alternate angles are equal, e.g., a = b, (ii) corresponding angles are equal, e.g., b = c, and (iii) interior angles on the same side of the transversal are supplementary, e.g., b + d = 2 right angles. b a d c bbreviation: (i) alt. angles between // lines; (ii) corr. angles between // lines; (iii) int. angles between // lines.
133 Theorem 8 (uclid I.32) (a) If one side of a triangle is extended, the exterior angle so formed is equal to the sum of the two interior opposite angles. (b) The angle sum of a triangle is equal to 2 right angles. bbreviation: (a) ext. angle of triangle; (b) angle sum of triangle. b a c 1c2 Theorem 9 (a) The sum of the interior angles of a convex polygon of n sides is 2n 4 right angles. (b) If the sides of a convex polygon are extended in order, the sum of the exterior angles so formed are 4 right angles. Theorem 10 (uclid I.26) (a) Two triangles are congruent if they have two pairs of equal angles, and the sides adjacent to the pairs of equal angles are equal: XY Z if = YXZ, = XY, = XY Z. X Y Z bbreviation: S (b) Two triangles are congruent if they have two pairs of equal angles, and the sides opposite to one pair if equal angles are equal: XY Z if = YXZ, = XY Z, = YZ. X Y Z bbreviation: S
134 Modern reorganization of uclid s ook I Theorem 11 (uclid I.5) If two sides of a triangle are equal, their opposite angles are equal; i.e., = = =. bbreviation: base angles, isos. triangle. Theorem 12 (uclid I.6) If two angles of a triangle are equal, their opposite sides are equal; i.e., = = =. bbreviation: sides opposite equal angles. Theorem 13 (uclid I.25) contrapositive (a) Two triangles are congruent if they have three pairs of equal sides: XY Z if X Z = XY, = YZ, = ZX. Y bbreviation: SSS Theorem 14 (a) Two triangles are congruent if they have two pairs of equal sides, and the opposite angles of one pair of equal sides are right angles: XY Z if Z Y X = YXZ =90, = YZ, = XY. bbreviation: RHS
135 asic onstruction 1 (uclid I.9) isect an angle. asic onstruction 2 (uclid I.10) The perpendicular bisector of a segment. asic onstruction 3 (uclid I.11) The perpendicular to a line at a point on the line. asic onstruction 4 (uclid I.12) The perpendicular to a line from a point outside the line. asic onstruction 5 (uclid I.23) Given a segment, to construct an angle equal to a given angle. asic onstruction 6 (uclid I.31) To construct the parallel to a line through a point not on the line. Theorem 15 (uclid I.34) (a) The opposite sides of a parallelogram are equal. (b) The opposite angles of a parallelogram are equal. (c) ach diagonal of a parallelogram bisects the area of the parallelogram. bbreviation: (a) opp. sides //gram; (b) opp. angles //gram. Theorem 16 The diagonals of a parallelogram bisect each other. bbreviation: diagonals //gram.
136 Modern reorganization of uclid s ook I Theorem 17 (uclid I.33) If one pair of opposite sides of a quadrilateral are equal and parallel, then the other pair of opposite sides are also equal and parallel. The quadrilateral is a parallelogram. bbreviation: 2 sides equal and //. Theorem 18 If the opposite angles of a quadrilateral are equal, then the quadrilateral is a parallelogram. bbreviation: opp. angles equal. Theorem 19 If the opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram. bbreviation: opposite sides equal. Theorem 20 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. P bbreviation: diagonals bisect each other. asic onstruction 7 (uclid I.46) To construct a square on a given segment.
137 Theorem 21 (uclid I.18) If two sides of a triangle are unequal, the greater side has the greater angle opposite to it. bbreviation: greater side opposite greater angle. Theorem 22 (uclid I.19) If two angles of a triangle are unequal, the greater angle has the greater side opposite to it. bbreviation: greater angle opposite greater side. Theorem 23 mong all lines drawn from a point to a line (not containing it), the perpendicular is the shortest. P Q Theorem 24 (uclid I.20) ny two sides of a triangle are together greater than the third side. bbreviation: triangle inequality
138 Modern reorganization of uclid s ook I Theorem 25 The line joining the midpoints of two sides of a triangle is parallel to the third side, and is equal to one half of its length. bbreviation: Midpoint theorem. Theorem 26 The line through the midpoint of a side of a triangle, parallel to a second side, passes through the midpoint of the third side. F P bbreviation: Intercept theorem. Theorem 27 If three or more parallel lines make equal intercepts on one traversal, then they make equal intercepts on any other transversal. bbreviation: Intercept theorem. asic onstruction 8 To divide a given segment into a given number of equal parts.
139 Theorem 28 (a) The three medians of a triangle are concurrent (at the centroid of the triangle). (b) The centroid of a triangle divides each median in the ratio 2:1. F G bbreviation: centroid theorem. H Theorem 29 (uclid IV.5) (a) ny point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. (b) ny point equidistant from the endpoints of a segment lies on the perpendicular bisector of the segment. bbreviation: perpendicular bisector locus. Theorem 30 The perpendicular bisectors of the three sides of a triangle are concurrent (at the circumcenter of the triangle). bbreviation: circumcenter theorem. O Theorem 31 The three altitudes of a triangle are concurrent (at the orthocenter of the triangle). Y bbreviation: orthocenter theorem. Z X H
140 Modern reorganization of uclid s ook I Theorem 32 (a) ny point on a bisector of an angle is equidistant from the lines containing the angle. (b) ny point equidistant from two intersecting lines lies on the bisector of an angle between the two lines. K P H bbreviation: angle bisector locus. Theorem 33 (uclid IV.4) The internal bisectors of the three angles of a triangle are concurrent (at the incenter of the triangle). bbreviation: incenter theorem. I
141 Theorem 35 The area of a rectangle is measured by the product of the measures of two adjacent sides. Theorem 36 The area of a triangle is equal to one half of the area of a rectangle on the same base and between the same parallels. Theorem 37 (uclid I.37) Triangles on the same base and between the same parallels are equal in area. Theorem 38 (uclid I.39) Triangles of equal area on the same base and on the same side of the base are between the same parallels. Theorem 39 (uclid I.41) If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangles is one half of that of the parallelogram. asic onstruction 9 (uclid I.45) To construct a triangle equal in area to a given quadrilateral.
142 Modern reorganization of uclid s ook I Theorem 40 (uclid I.47) In a right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the remaining two sides. Pythagorean theo- bbreviation: rem. K G H Theorem 41 (uclid I.48) If the area of the square on one side of a triangle is equal to the sum of the areas of the squares on the remaining two sides, then these two remaining sides contain a right angle. F L bbreviation: converse Pythagorean theorem. Y a c a b Z b X