Proposition 8.—This proposition is generally omitted. It is never used again by Euclid, and may be deduced from former propositions. Propositions 9, 10. — These propositions are two cases of the same proposition (see p. 138); as in Props. 5 and 6, the two cases are obtained by taking the point ( internal or external to AB. Props. 9 and 10 should be carefully compared with Props. 5 and 6. It will help the beginner if he remembers that the line AB is divided equally at P and unequally at Q in all the four propositions; that all four begin with AQ, QB, either as a rectangle or as squares; and that in Props. 9 and 10 there is no rectangle and no worů ' equal' in the enunciation. As in former propositions in this book, the proofs first given are Euclid's, and the Alternative proofs are obtained by deduction from previous propositions. The following points may be noticed in Euclid's proofs of Props. 9 and 10:(1) Both proofs are the same, the few words underlined in Prop. 10 constituting the only difference. (2) The proofs fall into two parts :-(a) dealing with the triangles APC, CED and DQB; and (b) the proof proper. (3) The proof proper begins with getting a value for the second side of the equation required, viz. twice the squares on AP, PQ. The gradual change of the squares into the square on AD, then into the squares on AQ, QD, and then into the squares on AQ, QB is one of the simplest and best examples of substitution to be found in Euclid. Of course these substitutions all really depend on Axiom 1. About the Alternative proofs it may be noticed that :(1) Props. 9 and 10 deal with squares, therefore the proofs depend on Props. 4 and 7, which also deal with squares. (2) The proof of Prop. 10 is the same as the proof of Prop. 9. Similar remarks about the use of these Alternative proofs might be made to the remarks already made about the Additional proofs of Props. 5 and 6. Proposition 11.—This is an important proposition, and is really one case of the following problem :-“ To divide a given straight line, either internally or externally, into two segments, so that the rectangle contained by the whole line and one segment may be equal to the square on the other segment.” The surd form of the roots of the quadratic equation a2 + ac a? O shows that the parts AH and HB are incommensurable, that is, the ratio of these parts cannot be expressed exactly by whole numbers, In Book VI. Euclid identifies dividing a line in medial section with dividing it in extreme and mean ratio. (See Euclid VI. 30.) The student should notice the numerous Exercises and Riders depending on this proposition and its figure. Propositions 12, 13. — These two propositions with I. 47 complete the discussion of the relation which exists between the square on one side of a triangle and the sum of the squares on the other two sides. The three results should be compared. They are often used, and in Trigonometry especially they are of great service. In the three figures of Prop. 13 the angle BCA is respectively acute, obtuse, and right. If we observe the convention about the sign of CD, noticing that it is negative in the first figure, positive in the second, and zero in the third, then, in all three cases, the square on BA is equal to the sum of the squares on BC, CA, together with twice the rectangle contained by BC and the projection of CA on BC. Proposition 14.—The algebraical equivalent of this proposition is x2 = ab, where ab denotes the area of the given rectilineal figure A. The proofs of Props. 11, 12, 13, 14 are riders on Props. 6, 4, 7, 5, in Book II. Prop. 11 depends on Prop. 6 for its proof ; Prop. 12 depends on Prop. 4; Prop. 13 depends on Prop. 7 ; and Prop. 14 depends on Prop. 5 for its proof. The last four propositions in Book II. therefore serve as excellent models for the use of Props. 4 to 7 in riders; and besides these propositions they all use Prop. 47 in Book I. SUMMARY OF THE RESULTS ARRIVED AT IN EUCLID, BOOK II. The enunciations in the second book of Euclid are so long and confusing that the student will find it better to learn the results of the propositions in the form given below. Prop. 8 is omitted as being unnecessary. The other thirteen propositions may be arranged in the following five groups :(1) One Theorem dealing with two lines AB and X: Prop. 1. X.AB = X.AC + X. CD + X.DB. (2) Four Theorems dealing with one line AB, divided into two parts AC, CB : Prop. 2. AB2 = AB.AC + AB.CB. = (3) Two Theorems, with two cases each, dealing with one line AB divided equally at P and unequally at (Q internal). ( external). Prop. 9. AQ2 + QB2 2(AP2 + PQ2) (Q internal). Prop. 10. AQ2 + QB2 = 2(AP2 + PQ2) (Q external). (4) Two Theorems dealing with triangles : Prop. 12. AB2 = BC2 + CA? + 2BC.CD. (BCA an obtuse angle). (5) Two Problems : Prop. 11. AB.HB AH. Prop. 14. EH2 = rectilineal figure A. The most important of all these are perhaps Props. 4, 5, 12, 13, and 11. ADDITIONAL PROPOSITIONS. 1. The square on any straight line is equal to four times the square on half the line. [Prove by II. 4.] 2. In a right-angled triangle, the square on the perpendicular drawn from the right angle to the hypotenuse is equal to the rectangle contained by the segments into which the perpendicular divides the hypotenuse. [I. 47 and II. 4.] 3. If a straight line be divided into two parts, the rectangle contained by the two parts is a maximum when the parts are equal. [11. 5.] 4. The difference of the squares on two sides of a triangle is equal to twice the rectangle contained by the base and the line intercepted between the middle point of the base and the foot of the perpendicular drawn from the vertex to the base. [By I. 47 and II. 5 (see No. 7, p. 139); or by II. 12 and 13.] 5. The sum of the squares on two sides of a triangle is equal to twice the square on half the base, together with twice the square on the line and 13.] 6. The sum of the squares on the diagonals of a parallelogram is equal to the sum of the squares on its four sides. [Use Nos. 5 and 1.] 7. The sum of the squares on the four sides of any quadrilateral is equal to the sum of the squares on its diagonals, together with four times the square on the line joining the middle points of the diagonals. [Use Nos. 5 and 1.] 8. The square on the straight line drawn from the vertex of an isosceles triangle to any point in the base is less than the square on either of the I. 47 ; or prove by No. 4.] 9. If A, B, C, D be four points taken in order along a straight line, prove that AB.CD + BC.AD + CA.BD O (Euler's Theorem). [By II. 1, observing the convention about signs.] 10. Produce BA to a point K, so that the rectangle AB, KB may be equal to the square on KA. equal to EB; and on AL, away from B, describe a square ALMK.] 11. If a straight line be divided internally in medial section, as in Prop. 11, and if a part be taken from the greater segment equal to the less, prove that the greater segment is divided in medial section. [II. 11 and 11. 3.] : ON THE ALGEBRAICAL SOLUTION OF GEOMETRICAL PROBLEMS. In the case of Prop. 11 it was shown that the problem could be stated algebraically thus, a (a – x) = x2; and in solving this quadratic equation we arrived at the following equation : ба? 4 The right-hand member of this equation might have suggested that in order to solve the problem we must find a line whose square is five times the square on half the given line. And this was done, for the square on EF is equal to five times the square on CE, which is equal to half AB. (See p. 145, Ex. 1. (vi).) Similarly in other cases a geometrical problem may often be stated algebraically, and the solution of an equation may suggest the geometrical solution of the problem. Take for example the following problem : To divide a given straight line into two parts, so that the square on one part may be equal to twice the square on the other part. Let a denote the length of the given straight line; and let a denote the length of one part; then a x will be the length of the other part. It is required to find the value of x which satisfies the equation. 2 (a X) 2 4ax (1) 2a + 12a (2) Now equation (1) suggests that we must find a line, the square on which is equal to twice the square on the given line; and equation (2) suggests that the required part x will be found by taking a line equal in length to twice the given line, and then adding or subtracting the length of the line suggested by equation (1). We know that the square on the diagonal of a square is equal to twice the given square; therefore we get the following solution :A P B с Q 2 - 2a ? 2 .. XC = Let AB be the given straight line ; Produce AB to C making AC = 2AB. such that AP2 2PB?, and AQ? 2BQ? E 2 |