Proposition 10. 295. If four quantities are in proportion, their like power's, or roots, are in proportion. Hyp. Let a:5=c:d. a" : 31 = c:an, To prove a с Proof Ō d an Сп ..ah : 31 = on:a. Extracting the nth root, 1 1 on dn Q.E.D. Proposition 11. 296. If any number of quantities are in proportion, any antecedent is to its consequent, as the sum of any number of the antecedents is to the sum of the corresponding consequents. Нур. Let a: b=c:d=e:f= etc. To prove a:b=c:d=etc. =a+ce+ etc. :b+d+f+etc. Proof: ab = ba, ad = bc, and af = be, etc., etc. (281) Adding, a(b + d + f + etc.) = b(a + ctet etc.). ..a:b=c:d=etc.=atote+etc. :b+d+f+etc. (283) Q. E. D. PROPORTIONAL LINES. 297. DEF. Two straight lines are said to be cut proportionally when the parts of one line are in the same ratio as the corresponding parts of the other line. Thus, AB and CD are cut pro P portionally at P and Q if AP : PB = CQ: QD. ó A B Proposition 12. Theorem. 298. A straight line parallel to one side of a triangle divides the other two sides proportionally. Hyp. Let DE be || to BC in the A ABC. H Н D E Proof. Take AH, any common measure of AD and DB, and suppose it to be contained 4 times in AD and 3 times in DB.. AD 4 (1) Through the several pts. of division of AD, DB draw || s to BC. They will divide AE into 4 equal parts and EC into 3 equal parts. If || 8 intercept equal lengths on any transversal, they intercept equal lengths on every transversal (152). AE 4 (2) 3 Therefore, from (1) and (2), AD AE (Ax. 1) DB EC• .:. AD: DB = AE: EC. CASE II. When A D and DB are incommensurable. Proof. In this case we know (232) that we may always find a line AG as nearly equal as we please to AD, and such that AG and GB are commensurable. Draw GH || to BC; then AG AH (Case I.) Now, these two ratios being always equal while the common measure is indefinitely diminished, they will be equal when G moves up to and as nearly as we please coincides with D. (233) AD AE Q.E.D. DB EC " 299. COR. 1. By composition (287), we have AD + DB :AD = AE + EC : AE, AB : AD = AC: AE. or 1. From a point E in the common base AB of two triangles ACB, ADB, straight lines are drawn parallel to AC, AD, meeting BC, BD at F, G: show that FG is parallel to CD. 2. In a triangle ABC the straight line DEF meets the sides BC, CA, AB at the points D, E, F respectively, and it makes equal angles with AB and AC: prove that BD: CD = BF: CE. Proposition 13. Theorem. 301. Conversely, if a straight line divides two sides of a triangle proportionally, it is parallel to the third side. Hyp. Let DE cut AB, AC in AB AC the A ABC so that AD AE which is impossible unless DE' coincides with DE. 302. DEF. When a finite straight line, as AB, is cut at a point X between A and B, it is said to be divided internally at X, X B and the two parts AX and BX are called segments. But if the straight line AB is produced, and cut at a point Y beyond AB, it is said to be divided externally at Y, and the parts AY and BY are called segments. The given line is the sum of two internal segments, or the difference of two external segments. |