« ForrigeFortsett »
1. Similar rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals.
2. Two magnitudes are said to be reciprocally proportional to two others, when one of the first pair is to one of the second, as the remaining one of the second is to the remaining one of the first.
3. A straight line is said to be cut in extreme and mean ratio, when the whole is to one of the segments, as that segment is to the other.
4. The altitude of any figure is the straight line drawn from its vertex perpendicular to its base.
5. A straight line is said to be cut harmonically, when it is divided into three segments, such that the whole line is to one of the extreme segments, as the other extreme segment is to the middle one.
Triangles and parallelograms of the same altitude are one to another as their bases.
Let the triangles ABC, ACD, and the parallelograms EC, CF have the same altitude, viz., the perpendicular drawn from the point A to BD : then, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF.
Produce BD both ways, and take any number of straight lines BG, GH, each equal to BC; and any number DK, KL, each equal to CD; and join AG, AH, AK, AL. Then, because. CB, BG, GH are all equal, the triangles ABC, AGB, AHG are (I. 38.) all equal. Therefore, whatever multiple the base HC is
of BC, the same multiple is the triangle AHC of ABC. For the same reason, whatever multiple LC is of CD, the same multiple is the triangle ALC of ADC. Also, if the base HC be equal to CL, the triangle AHC is equal (I. 38.) to ALC; and if the base HC be greater than CL, likewise (1. 38. cor.) the triangle AHC is greater than ALC; and if less, less. Therefore, since there are four magnitudes, viz., the two bases BC, CD, and the two triangles ABC, ACD; and of the base BC, and the triangle ABC, the first and third, any like multiples whatever have been taken, viz., the base HC, and the triangle AHC; and of the base CD, and the triangle ACD, the second and fourth, have been taken any like multiples whatever, viz., the base CL, and the triangle ALC; and that it has been shown, that, if the base HC be greater than CL, the triangle AHC is greater than ALC; if equal, equal; and if less, less : therefore (V. def. 5.) as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.
Again, because (I. 41.) the parallelogram CE is double of the triangle ABC, and the parallelogram CF of the triangle ACD, and that (V. 15.) magnitudes have the same ratio which their like multiples have; as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF. But it has been shown, that BC is to CD, as 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 (V. 11.) as the base BC is to the base CD, so is the parallelogram E to the parallelogram CF. Wherefore, triangles and parallelograms, &c.
Cor. 1. From this it is plain, that triangles and parallelograms which 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 (I. 33.) to that in which their bases are, because the perpendiculars are both equal and parallel to one another. Then, if the same construction be made as in the proposition, the demonstration will be the same.
Cor. 2. Hence, if A, B, C be any three straight lines, we have A:B:: A.C:B.C.
Cor. 3. So likewise, if the straight lines A, B, C, D be proportional, and E and F be any other straight lines, we shall have, according to the preceding corollary, and the 11th proposition of the fifth book, A.E:B.E:: C.F: D.F.
PROP. II. THEOR. If a straight line be parallel to the base of a triangle, it cuts the other sides, or those produced, proportionally, and the segments between the parallel and the base are homologous to one another : and (2.) if the sides of a triangle, or the sides produced, be cut proportionally, so that the seginents between the points of section and the base are homologous to one another, the straight line which joins the points of section is parallel to the base.*
1. Let DE be parallel to BC, one of the sides of the triangle ABC; BD: DA::CE: EA.
Join BE, CD. Then (I. 37.) the triangles BDE, CDE are equal, because they are on the same base DE, and between the same parallels, DE, BC. Now ADE is another triangle, and (V. 7.) equal magnitudes have to
the same, the same ratio ; therefore, as the triangle BDE to ADE, so is the triangle CDE to ADE. But
(VI. 1.) as the triangle BDE to ADE, so is BD to DA; because, having the same altitude, viz., the perpendicular drawn from E to AB, they are to one another as their bases ; and for the same reason,
as the triangle CDE to ADE, so is CE, to EA. Therefore (V. 11.) as BD : DA:: CE: EA.
2. Next, let the sides AB, AC of the triangle ABC, or those produced, be cut proportionally in the points D, E; that is, so that BD:DA:: CE : EA, and join DE; DE is parallel to BC. The same construction being made, because (hyp.)
as BD: DA :: CE : EA; and (VI. 1.) as BD to DA, so is the triangle BDE to the triangle ADE; and
* The enunciation of this proposition, as given by Dr. Simson, is as follows : “ If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those 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.” This enunciation is defective, and might lead to error in its application, as it does no point out what lines are homologous to one another in the analogy.
It is plain that instead of one proposition, this is in reality two, which are converses of one another. The same is the case in the third, fourteenth, fifteenth, sixteenth, seventeenth, and twenty-second of this book.
as CE to EA, so is the triangle CDE to ADE: therefore (V. 11.) the triangle BDE is to ADE, as the triangle CDE to ADE; that is, the triangles BDE, CDE have the same ratio to ADE; and therefore (V. 9.) the triangles BDE, CDE are equal, and they are on the same base DE, and on the same side of it: therefore (I. 39.) DE is parallel to BC. Wherefore, if a straight line, &c.
Cor. The triangles which two intersecting straight lines form with two parallel ones, have their sides which are on the intersecting lines proportional ; and those sides are homologous which are in the same straight line: and, (2.) conversely, if two straight lines form with two intersecting ones triangles which have their sides that are on the intersecting lines proportional, the sides which are in the same straight line with one another being homologous, those straight lines are parallel.
Let DE and BC (first and third figures) be the parallels, and let them be cut by the straight lines BD, CE, which intersect each other in A; then BA : AC : : DA : AE. For, since BD: DA : : CE : EA, we have, by composition in the first figure, and by division in the third, BA : DA:: CA : EA, and, alternately, BA : AC :: DA : AE.
2. But if BA : AC :: DA : AE, DE and BC are parallel. For, alternately, BA: DA ::CA: EA; then, in the first figure by division, and in the third by composition, we have BD: DA:: CE: EA; and therefore, by the second part of this proposition, DE is parallel to BC.
PROP. III. THEOR.
The straight line which bisects an angle of a triangle, divides the base into segments which have the same ratio to one another as the adjacent sides of the triangle have: and (2.) if the segments of the base have the same ratio as the adjacent sides, the straight line drawn from the vertex to the point of section, bisects the vertical angle.
1. Let the angle BAC of the triangle ABC be bisected by the straight line AD; BD: DC :: BA : AC.
Through C draw (1. 31.) CE parallel to DA: then (I. 29. cor. 1.) BA produced will meet CE ; let them meet in E. Because AC meets the parallels AD, EC, the angle ACE is equal (I. 29.) to the alternate angle CAD; and because BAE meets the same parallels, the angle E is equal (I. 29. part. 2.) to BAD: therefore (I. ax. 1.)
the angles ACE, AEC are equal, because they are respectively equal to the equal angles, DAC, DAB; and consequently AE is equal (I. 6.) to AC. Now (VI. 2.) because AD is parallel to EC, one of the sides of the triangle BCE, BD : DC :: BA : AE; but AE is equal to AC; therefore (V. 7.) BD:DC:: BA: AC.
2. Let now BD:DC::BA: AC, and join AD; the angle BAC is bisected by AD. The same construction being made, because
(hyp.) BD:DC:: BA: AC; and (VI. 2.) BD: DC::BA: AE, since AD'is parallel to EC: therefore (V. 11.) BA : AC :: BA: AE: consequently (V. 9.) AC is equal to AE; and (I. 5.) the angles AEC, ACE are therefore equal. But (I. 29.) the angle BAD is equal to E, and DAC to ACE; wherefore also BAD is equal (I. ax. ).) to DAC; and therefore the angle BAC is bisected by AD. The straight line, therefore, &c.
PROP. A. THEOR. * If an exterior angle of a triangle be bisected by a straight line which also cuts the base produced, † the segments between the bisecting line and the extremities of the base, have the same ratio to one another, as the other sides of the triangle have: and (2.) if the segments of the base produced, have the same ratio which the other sides of the triangle have, the straight line drawn from the vertex to the point of section bisects the exterior angle of the triangle.
1. Let the exterior angle CAE of any triangle ABC be bisected by AD which meets the base produced in D; BD: DC:: BA: AC.
Through C draw (1. 31.) CF parallel to AD, and because AC meets the parallels AD, FC, the angle ACF is equal (I. 29.) to
This proposition was inserted in the Elements by Dr. Simson ; who, besides stating that it is merely a second case of the third proposition, remarks, that “the demonstration of it is very like to that of the first case, and upon this account may, probably, have been left out, as also the enunciation, by some unskilful editor. Dr. Elrington conceives this view to be corroborated by the expression, “which also cuts the base,” in the enunciation; as this expression is necessary only in relation to the line which bisects the exterior angle. The enunciations of both cases may be easily combined in one. For the sake of simplicity, however, they are here allowed to continue separate.
+ If the triangle be isosceles, the line bisecting the exterior angle at the vertex is parallel to the base. In this case, the segments may be regarded as ipfiuite, and therefore equal, their difference, the base being infinitely small in comparison of them.