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taken, viz. the base HC and triangle AHC; and of the base CD and triangle ACD, the second and fourth, have been taken any equimultiples whatever, viz. the base CL and triangle ALC; and that it has been shown, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; and if equal, equal; and if less, less: therefore,* * 5 Def. as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.
And because the parallelogram CE is double of the triangle ABC, and the parallelogram CF double of * 41. 1. the triangle ACD, and that magnitudes have the same ratio which their equimultiples have; as the triangle * 15. 5. ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF: and because it has been shown, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is the parallelogram * 11. 5. EC to the parallelogram CF. Wherefore triangles, &c. Q. E. D.
COR. From this it is plain, that triangles and parallelograms that have equal altitudes, are one to another as their bases.
Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are, because the perpendiculars are both equal and parallel to one another. * 28. 1 Then, if the same construction be made as in the proposition, the demonstration will be the same.
PROP. II. THEOR.
If a straight line be drawn parallel to one of the sides
* 37. 1.
of a triangle, it shall cut the other sides, or these produced, proportionally.
And if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle.
Let DE be drawn parallel to BC, one of the sides of the triangle ABC: then BD shall be to DA, as CE to EA.
Join BE, CD; then the triangle BDE is equal* to the triangle CDE, because they are on the same base DE, and between the same parallels DE, BC: but ADE is another triangle, and equal magnitudes have the same ratio to the same magnitude; therefore, as the triangle BDE to the triangle ADE, so is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, so is* BD to DA, because having the same altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the triangle CDE to the triangle ADE, so is CE to EA: therefore, as 11. 5. BD to DA, so is CE to EA.*
Next, let the sides AB, AC of the triangle ABC, or
A A A
these produced, be cut proportionally in the points D, E, that is, so that BD be to DA, as CE to EA, and join DE; DE shall be parallel to BC.
The same construction being made, because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE;
and as CE to
EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as * 11. 5. the triangle CDE to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle ADE; and therefore the triangle BDE is * 9. 5. equal to the triangle CDE: and they are on the same base DE; but equal triangles on the same base, and on the same side of it, are between the same parallels;* * 39. 1 therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D.
PROP. III. THEOR.
If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another.
And if the segments of the base have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section, divides the vertical angle into two equal angles.
Let ABC be any triangle, and let the angle BAC be divided into two equal angles by the straight line AD: BD shall be to DC, as BA to AC.
Through the point C draw CE parallel to DA, and • 31. 1. let BA produced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD: but the angle * 29. 1. CAD, by the hypothesis, is equal to the angle BAD; wherefore the angle BAD is equal* to the angle ACE. * 1 Ax. Again, because the straight
line BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and opposite angle AEC: but
*Const. multiples* whatever, of A, C, and G, H any equimultiples whatever of B, D: therefore A is to B, as C is 5 Def. to D.*,
Next, let A the first be the same part of B the second, that C the third is of D the fourth: A shall be to B, as C is to D. For since A is the same part of B that C is of D; therefore B is the same multiple of A, that D is of C: wherefore, by the pre*B. 5. ceding case, B is to A, as D is to C; and inversely, A is to B, as C is to D. Therefore, if the first be the same multiple, &c. Q. E. D.
Δ B C D
If the first be to the second as the third to the fourth, and if the first be a multiple, or part of the second; the third is the same multiple, or the same part of the fourth.
Let A be to B, as C is to D; and first let A be a multiple of B; C shall be the same multiple of D. Take E equal to A, and whatever multiple A or E is of B, make F the same multiple of D: then, because* A is to B, as C is to D; and of B the second, and D the fourth equimultiples E and F have been taken; therefore *Cor. 4. A is to E, as C to F:* but A is equal to E, therefore C is equal to F:* but
* A. 5.
* Const. F is the same* multiple of D, that A is
See the figure at the top of this
of B: therefore C is the same multiple of D, that A is of B.
Next, let A the first be a part of B the second; C the third shall be the same part of D the fourth. Because A is to B, as C is to D; then, inversely, B is* to A, as D to C: but A is a part of B, therefore B