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have the same ratio, viz. A: B::E:F, and B: C:: D:E, then A: C::D: F.

Take of A, B, and D, any equimultiples mA, mB, mD; and of C, E, F, any equimultiples nC, nE, nF.

Because A: B::E: F, and because also A: B:: mA: mB, and E:F::nE: nF (v. Prop. 15); therefore, mA: mB:: nE: nF (v. Prop. 11). Again, because B:C::D: E, mB:nC::mD:nE (v. Prop. 4); and it has been just shown that mA: mB::nE: nF; therefore, if mA 7 nC, mD7nF (v. Prop. 21); _if_mA=nC, mD=nF; and if mAnC, mDnF. Now, mA and mD are any equimultiples of A and D, and nC, nF, any equimultiples of C and F; therefore, A: C:: D: F (v. Def. 5).

Next, let there be four magnitudes, A, B, C, and D, and other four, E, F, G, and H, which, taken two and two, in a cross order, have the same ratio, viz. A:B::G: H; B:C:: F: G, and C: D:: E: F, then A: D:: E: H. For, since A, B, C are three magnitudes, and F, G, H other three, which, taken two and two, in a cross order, have the same ratio, by the first case, A:C::F: H. But C: D:: E: F, therefore again, by the first case, A: D:: E: H. In the same manner may the demonstration be extended to any number of magnitudes.


If the first has to the second the same ratio which the third has to the fourth; and the fifth to the second, the same ratio which the sixth has to the fourth; the first and fifth, together, shall have to the second, the same ratio which the third and sixth, together, have to the fourth.

Let A: B::C: D, and also E:B::F: D, then A+E: B::C+F: D.

Because E: B:: F: D, by inversion, B: E::D: F. But by hypothesis, A: B:: C: D, therefore, ex æquali (v. Prop. 22), A:E::C: F, and, by composition, A+E : E::C+F:F (v. Prop. 18). And again, E:B::F:D (Hyp.), therefore, ex æquali, A+E:B::C+F: D.


If four magnitudes be proportionals, the sum of the first two is to their difference as the sum of the other two to their difference.

Let A B C D; then if A7B, A+B: A-B::C+D :C-D; or if AB, A+B:B-A::C+D: D-C. For, if A7B, then, because A: B:: C: D, by division (v. Prop. 17), A-BB:: C-D: D, and by inversion (v. Prop. A), B: A B::D:C-D. But, by composition (v. Prop. 18), A+B:B:: C+D: D, therefore, ex æquali (v. Prop. 22), A+B: A−B:: C+D:C-D. In the same manner, if B7A, it is proved, that A+B B-A::C+D:D-C.

Note. For Remarks, see APPENDIX E.



1. Rectilinear figures are said to be similar, when they have their several angles equal, each to each, and the sides about the equal angles proportional.

2. Two sides of one figure are said to be reciprocally proportional to two sides of another figure, when one of the sides of the first is to one of the sides of the second, as the remaining side of the second is to the remaining side of the first.

3. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.

4. The altitude of any figure is the straight line drawn from its vertex perpendicular to its




Triangles (CAB, EAD) and parallelograms (CKAB, EALD), having the same altitude, are to one another as their bases (CB, ED).

Since the triangles and parallelograms have the same altitude AP, their bases must lie in the same straight line CD. Produce CD both ways towards H and N; and in the



line produced from BC, take consecutively any number of parts, as CF, FG, GH, each equal to BC; and also in the line produced from ED, take any number of parts, as DM, MN, each equal to ED; and join AF, AG,


AH, AM, and AN. Then the tri- H G F C B PE angles EAD, DAM, MAN are all equal (i. Prop. 38); and consequently the triangle EAN is the same multiple of EAD, as the base EN is of ED. For like reasons, the triangle BAH is the same multiple of BAC as the base BH is of BC. But if the base EN=BH, then likewise the triangle EAN = BAH (i. Prop. 38); and consequently also, if EN BH, EAN BAH; and if EN ▼BH, EAN▼BAH. The same conclusion may be drawn, whatever multiples be assumed of the bases CB and ED; and therefore CAB: EAD::CB:ED (v. Def. 5).

The parallelograms CKAB, EALD are double of the triangles CAB, EAD (i. Prop. 41), and therefore, being equimultiples of them, are proportional to them (v. Prop. 15), or CKAB: EALD:: CAB: EAD; but CAB: EAD:: CB: ED; and therefore CKAB: EALD:: CB: ED (v. Prop. 11).

COR.-Hence it is manifest that triangles and parallelograms, having equal altitudes, are proportional to their bases. For, if the figures be so placed that their bases may lie in the same straight line, then since their altitudes are equal, the straight line joining the vertices of the triangles will be parallel to that in which their bases lie (i. Prop. 33), and the demonstration proceeds as above.


If a straight (DE) be drawn parallel to one of the sides (BC) of a triangle (BAC), it will cut the other sides, or the other sides produced, proportionally: and if the sides (AC, AB), or the sides produced, be cut proportionally, the straight line (DE) which joins the points of section, will be parallel to the remaining side of the triangle.

Join BE, CD; and since DE is parallel to BC (Hyp.), the triangles BED, CDE are equal (i. Prop. 37); and therefore

have the same ratio to the triangle DAE (v. Prop. 7),
:: BD : DA, and also CDE: DAE::CE: EA
(vi. Prop. 1); therefore BD: DA::CE: EA (v. Prop. 11), E
or the sides of the triangle are cut proportionally
by the line drawn parallel to the base.

But if the hypothesis be that the sides of the triangle, or those sides produced, are cut proportionally, then the line joining the points of section must be parallel to the base. For the



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triangle BED : DAE :: BD : DA, and also CDE: DAE::CE: EA (vi. Prop. 1); but BD:DA ::CE: EA (Hyp.), and therefore BED: DAE B :: CDE: DAE (v. Prop. 11), and consequently BED=CDE (v. Prop. 9); but these equal triangles being upon the same base DE, must be between the same parallels (i. Prop. 39), and therefore DE joining the points of section of the sides or of the sides produced, is parallel to the base BC.


If an angle (BAC) of a triangle be bisected by a straight line (AD) which also cuts the base, the segments of the base shall be proportional to the other sides of the triangle (BD: DC:: BA: AC); or, if the segments of the base be proportional to the other sides of the triangle, the straight line (AD) drawn from the vertex to the point of section, shall bisect the vertical angle.


From the point C draw CE parallel to DA and meeting BA produced to E. Then, because DA and CE are parallel, ▲ DAC = ≤ ACE, and also / DAB = AEC (i. Prop. 29); but DAC = ≤ DAB (Hyp.), and therefore also ACE = ▲ AEC, and AE = AC (i. Prop. 6). But since DA is parallel to CE, BD: DC :: BA: AE (vi. Prop. 2); and since AE = AC, BA AC :: BA : AE (v. Prop. 7); therefore BD: DC ::BA: AC, or the segments of the base made by the line bisecting the opposite angle, are proportional to the sides containing that angle.



If the hypothesis be that BD:DC:: BA: AC, then AD

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