(fig. 36) be divided into equal parts, and if lines be drawn F36 through the points of division parallel to the base P R, then A R will be divided into the same number of equal parts as A P; in other words A F=F G=G H=H I, &c. Through the points F, G, H, &c., draw the lines F K, G L, H M, &c., parallel to A P. Then the triangles A B F, FK G, GLH, &c. are equal. Why? Because they all have a side and two adjacent angles equal (55): namely,* A B=F K=G L, &c., because parallels comprehended between parallels are equal (38); the angle B A F=K F G=LG H, &c., because internal external angles are equal (34); and lastly A B F=F K G=G L H, &c., because angles which have their sides parallel and directed the same way are equal (39). Therefore A F=F G-G H, &c., . and AR is divided into the same number of equal parts as A P. Secondly we say that if A P be to any portion A E (fig. 36) as two whole numbers, and E I be drawn par- F 36 allel to P R, then A R will be to A I in the ratio of the same two numbers, and the two sides will be divided proportionally. For if A P be divided, for example, into 7 equal parts, and A E contain 4 of them, then their ratio can be expressed by whole numbers, and we have A P: A E: 7:4. But by the preceding proposition A R is also divided into 7 equal parts, and AI contains 4 of them, wherefore AR:AI::7: 4. Then leaving out the ratio 7:4 common to the two proportions (64), we have A P: A E::AR: AI. Thus the second part of the proposition is demonstrated.. Thirdly we say that even when the ratio of A B to A D (fig. 37) cannot be expressed by whole F37 numbers, still, if D E be drawn parallel to B C, we have universally the proportion A B: A D::A C:A E, or inverting the means (65) A B:A C::AD: A E. The method of proof is by what is called a reductio ad absurdum, and is as follows. If the fourth term of the above proportion be not A E, it must be some line either greater or less than A E. Now if we can show that it is absurd to sup pose it either greater or less than A E, the other terms remaining the same, then the fourth term must be A E, and the proportion will be true. Let us then in the first place take a fourth term less than A E, for instance A Q, Then the proportion will be A B A C A DA O. Now suppose A C divided into any number of equal parts each less than O E. One point of division must fall between 0 and

E. Let that point be G and draw F G parallel to B C. Then, since AC is to A G in the ratio of two whole numbers, we have, according to the preceding demonstration, AB: AC::AF: A G. But by hypothesis we had A B:A C::AD: A O. These two proportions have the ratio A B A C common. Leaving it out (64), we have the proportion A F:A G::A D:A O. Inverting the means (65) we have A F: A D::AG: A 0, which is manifestly absurd, since it asserts that a less is to a greater as a greater is to a less. This absurdity arises from supposing that the fourth term could be less than A E. Therefore we know that it cannot be less. If now we should suppose it greater, and take O on the other side of E, by repeating the same course of reasoning verbatim, we should arrive at a similar absurdity. The fourth term then cannot be greater than A E. And since it can be neither less nor greater than A E, it must be A E, and the proportion above enunciated is rigorously true, namely A B AC::AD: A E or A B A D::AC:A E. From this proportion, we have by subtraction (68) A B-A D: A C-A E::A D: A E. But A B-A D=B D, and A C-A E-C E. Therefore B D:CE::AD: A E, or inverting the means BD:AD::CE: A E.

71. The converse of the above proposition is equally true, namely-If a line be drawn so as to divide two sides of a triangle proportionally, that line is parallel to the third F38 side. Thus if D E (fig. 38) be so drawn that we have the proportion BD:DA::CE: E A, then we say that DE is parallel to B C. For if it is not, some other line drawn through D must be. Suppose that line to be D F. Then if D F is parallel to B C, we have by the preceding proposition B D:DA::CF: F A. But by the conditions of the proposition we had B D: DA::CE: CA. From these two we have (64) C F: FA:CE: EA. This proportion is absurd, as will be seen by inverting the means. For then we have C F:CE::FA: E A, that is a greater is to a less as a less is to a greater. Therefore no line different from D E, can be drawn through D parallel to B C.

72. We shall now solve several problems, which depend upon the proposition of article 70. We begin with the following-To find a fourth proportional to three given F 39 lines. Let the three lines be A, B, C (fig. 39). Draw

the indefinite lines D P, D R, making any angle at pleasure. Take D E A, D F B, and D G=C. Join E and F, and through G draw G H parallel to E F. Then D H will be the fourth term required. For we have the proportion DE: DF:: DG:DH. This geometrical operation corresponds to the Single Rule of Three in arithmetic.

73. To divide a given straight line into any number of equal parts. Suppose it were required to divide A B (fig. 40) into six equal parts. Draw the line AP indefi- F 40 nitely. Take A C of any convenient length, and apply it six times to A P. Through H, the last point of division, draw H B. Through C draw C I parallel to H B. AI 'will be a sixth part of A B. For AI: A B:: AC: A H. But by construction A C is one sixth of A H. Therefore A I is one sixth of A B. Apply A I six times to A B, and A B will be divided into six equal parts.

74. -To divide a given line into parts proportional to any given lines. Suppose it were required to divide D F (fig. 41) into three parts proportional to the three given F 41 lines A, B, C. Draw D P indefinitely. Take D G=A, G H=B, H L=C. Draw L F, and through H and G draw H K and G I parallel to L F. Then D I:IK::D G: G H or A: B. Also DK:IK:: DH: G H and D K: KF::DH: H L. Inverting the means in the two last and leaving out the common ratio D K: D H, we have IK:KF:: GH:H L or B: C. Hence D I: A::I K: B:: KFC. Therefore the line D F is divided as required.

75. -To divide one side of a triangle into two parts proportional to the other two sides. Suppose it were required to divide B C (fig. 42) into two parts proportional to A F42 B and A C. Draw A D so as to bisect the angle B A C, and D will be the point of division. For we shall have the proportion C D:DB::C A: A B. Why?-Draw B E parallel to A D till it meets C A produced. Then C D : D B::CA:A E. But A E=A B. For the angle BE A=D A C (34) and E B A B A D. Now B A D=D A C by construction. Therefore A E B A B E. Then (50) À E=A B. Substituting A B for A E in the last proportion, we have C D: D B:: CA: A B.

76. Through a given point in an angle, to draw a line so that the parts intercepted between the point and the sides of the angle shall be equal. Suppose it were required to

F43 draw through the point B in the angle D A E (fig. 43) a line D E in such a manner that D B should be equal to B E. Draw B C parallel to A E. Take C D C A, and through the points D, B, draw D E. Then D B=B E, for D B: BE::DC: CA; and D C=C A.

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Similar Triangles.

77. Two triangles are said to be similar when they are equiangular with respect to each other—. Thus if A B F44 C (fig. 44)=C D E, B A C D C E, and A C B C E D, then the triangles A B C and C D E are similar. Now there are three cases in which two triangles are equiangular. 1. -When they have their sides parallel each to each (39). 2. When they have their sides perpendicular each to each. For then by turning one of the triangles about one of its vertices by the space of a quadrant, the sides will become parallel each to each. 3. When they have an angle of the one equal to an angle of the other, and the sides including these angles proportional-. Thus if the F45 angle A=A (fig. 45), and if A B: A D:: AC: AF, then we say the triangles are equiangular. For if the side A D be placed upon A B, since the angles at A are equal A F will fall upon A C. Then, from the above proportion, D F must be parallel to B C (71). Consequently the angle A D F=A B C and A F D=A C B (34), and the two triangles are equiangular. In the above three cases, then, according to the definition, two triangles are similar.

78. Two similar triangles have their homologous sides proportional By homologous sides we mean those which have corresponding positions with respect to the equal angles. Thus in the similar triangles BAC and DC F44 E (fig. 44) A B is homologous to C D, being opposite to equal angles, and so of the rest. We are now to demonstrate the following proportion A B: CD:: BC: DE:: A CC E. Let the two triangles be so placed that A E and CE shall be in the same straight line. Produce A B and

ED till they meet in F. Now B C is parallel to E F, because the angle B C A=DEC (34). Also CD is parallel to A F, because the angle D C E=B A C. Then we have (70) A C:CE::AB: BF. But B F C D (38). Therefore AC:CE::A B:C D. Again A C:CE::FD; D E. But F D B C. Therefore A

C:CE: BC: D E. Thus the three ratios formed by the three couples of homologous sides are equal, and give the continued proportion A B C D::BC:D E:: A C: CE.

79. The converse of the above proposition is equally true, namely-If two triangles have their homologous sides proportional, they are similar. If the triangles A B C and DEF (fig. 46) give the proportion A B D E::BC: E F 46 F:: AC: D F, then we say they are similar. Draw D G so as to make the angle G D F=A. Draw F G so as to make the angle D F G=C. Then the angle G=C (48), and the triangles A B C and D G F are similar by construction. Now if we prove that the triangle E D F is equal to D G F, it will follow that E D F is similar to A BC. By the conditions we have A C D F :: A B : DE, and by construction we have (73) A C: D F :: A B : D G. In these proportions the three first terms are the same, and therefore (63) the fourth terms must be equal. Thus DE=D G. Again by the conditions we have A C: D F:: BC: E F, and by construction, A C: DF::BC:F G. Therefore E F=F G. Then the triangles EDF and D F G are equal (57) having their three sides respectively equal. Consequently E D F is similar to A B C, which was to be demonstrated.

80. If from any point in a semicircumference, a line be drawn perpendicular to the diameter, it will be a mean proportional between the two segments of the diameter—. Thus we say that A D (fig. 47) is a mean proportional between B F47 D and D C, or in other words that B D: AD::AD:D C. Draw the two chords A B and A C. Then the two triangles A B D and A D C will be similar. For the angle A D B A D C, being right angles. Also since B A C is a right angle (42), B A D is at the same time a complement of DA C and D B A. Therefore (21) DA C=D B A, and as the third angles must be equal (48), the two triangles are similar. We have then B D homologous to A D, and A D homologous to D C. Therefore BD:AD::AD: D C. If then it be required to find a mean proportional between two given lines, we have only to make these two lines a diameter, and at the point of junction erect a perpendicular to meet the circumference. This we shall find to be a very useful problem.

81.

If from a point without a circle a tangent and secant