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triangles A BE, FGL are equiangular, EB: BA :: LG:GF; and also, (VI. def. 1.) because the polygons are similar, AB : BC :: FG : GH; therefore, ex æquo, (V. 22.) EB : BC :: LG: GH; that is, the sides about the equal angles EBC, LGH are proportionals; therefore (VI. 6.) the triangles EBC, LGH are equiangular, and (VI. 4. and def. 1.) similar. In the same manner, the triangles ECD, LHK may be shown to be similar. Therefore the similar polygons AD, FK are divided into the same number of similar triangles.

2. Because the triangles ABE, FGL are similar, ABE has to FGL the duplicate ratio (VI. 19.) of that which the side BE has to GL. For the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL.

Therefore (V. 11.) as the triangle A BE to FGL, so is the triangle BEC to GLH. Again, be. cause the triangles EBC, LGH are similar, EBC has to LGH the duplicate ratio of that which EC has to LH. For the same reason, the triangle ECD has to LHK the duplicate ratio of that which EC has to LH. As, therefore, (V. 11.) the triangle EBC is to LGH, so is the triangle ECD to LHK. But it has been proved that the triangle EBC is to LGH, as the triangle ABE to FGL. Therefore (V. 11.) as the triangle ABE is to FGL, so is the triangle EBC to LGH, and ECD to LHK: and therefore (V. 12.) as one of the antecedents to its consequent, so are all the antecedents to all the consequents. Wherefore as the triangle ABE is to FGL, so is the polygon AD to the polygon FK.

3. The triangle ABE has (VI. 19.) to FGL, the duplicate ratio of that which the side AB has to the homologous side FG. Therefore, also, (V. 11.) the polygon AD has to the polygon FK the duplicate ratio of that which AB has to the homologous side FG. Similar polygons, therefore, &c.

Cor. 1. In like manner, it may be proved, that similar figures of four sides, or of any number of sides, are one to another in the duplicate ratio of their homologous sides, and the same has already been proved (VI. 19.) in respect to triangles. Therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides.

Cor. 2. If to AB, FG, two of the homologous sides, a third proportional M be taken, AB has (V. def. 11.) to M the duplicate ratio of that which AB has to FG. But the four-sided figure or polygon upon AB has likewise to the four-sided figure or polygon upon FG the duplicate ratio of that which AB has to FG. Therefore as AB is to M, so is the figure upon AB to the figure upon FG; which was also proved (VI. 19. cor.) in triangles. Therefore, universally, if three straight lines be proportionals, the first is to the third, as any rectilineal figure upon the first, to a similar and similarly described figure upon the second.

Cor. 3. Because all squares are similar figures, the ratio of any two squares to one another, is the same as the duplicate ratio of their sides ; and bence, also, any two similar rectilineal figures are to one another, (V. 11.) as the squares described on their homologous sides. *

Cor. 4. The perimeters of the figures, that is, the sums of their sides are proportional to the homologous sides. For (hyp. and VI. def. 1.) the sides are proportional, and therefore, alternately, AB : FG:: BC: GH::CD : HK::DE: KL::EA : LF: wherefore (V. 12.) AB : FG :: AB + BC + CD + DE + EA: FG + GH + HK + KL + LF.

PROP. XXI. THEOR. RECTILINEAL figures which are similar to the same figure, are similar to one another.

Let each of the rectilineal figures, A, B, be similar to C: A is similar to B.

Because A is similar to C, they are (VI. def. 1.) equiangular, and have also their sides about the equal angles proportional. For the same reason, B and C are equiangular, and have their sides about the equal angles proportional. Therefore the figures A, B are each of them equiangular to C, and have the sides about the equal angles of each of them, and of C proportional : wherefore (I. ax. 1.) A and B are equiangular; and (V. 11.) they have their sides about the equal angles proportionals. Therefore (Vl. def. 1.) A is similar to B. Rectilineal figures, therefore, &c.

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PROP. XXII. THEOR. If four straight lines be proportionals, the similar rectilineal figures, similarly described upon them, are also proportionals : and (2.) if the similar rectilineal figures, similarly described upon four straight lines, be proportionals, those lines are proportionals.

Let AB : CD:: EF: GH; and upon AB, CD let the similar

* This corollary and the preceding afford easy means of comparing similar rectilineal figures. Thus, the second enables us to find two lines having the same ratio as the figures, and the third affords, in general, the easiest means of exhibiting their ratio in numbers, or by algebraic symbols ; the area of a square (I. 46. cor. 5.) being found simply by multiplying a side by itself. Thus, if AB were 3 inches, and FG 2 inches, the areas of the squares of these lines would be 9 square inches and 4 square inches ; and therefore the figure AD would be to FK as 9 to 4, or would be rather more than twice as great. If, again, AB were 5 feet, and FG 3 feet, AD would be to FK as 25 to 9, or would be nearly three times as great.

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rectilineal figures KAB, LCD be similarly described; and upon EF, GH the similar rectilineal figures MF, NH in like manner : KAB is to LCD, as MF to NH.

To AB, CD take (VI. 11.) a third proportional X; and to EF, GH a third proportional O. Then, because

AB: CD :: EF: GH, and (V. 11.) CD:X :: GH: 0; therefore, ex æquo, as AB :X :: EF:0. But (VI. 20. cor. 2.) as AB to X, so is the figure KAB to LCD; and as EF to 0, so is MF to NH: therefore (V. 11.) as KAB to LCD, so is MF to NH.

Again, if the rectilineal figure KAB be to LCD, as MF to NH; AB: CD:: EF: GH.

To AB, CD, and EF, take (VI. 12.) a fourth proportional PR, and on it describe (VI. 18.) the figure SR similar and similarly situated to either of the figures MF, NH. Then, because AB:CD:: EF: PR, and that upon AB, CD are described the similar and similarly situated rectilineal figures KAB, LCD, and upon EF, PR, in like manner, the similar rectilineal figures MF, SR; KAB: LCD :: MF: SR; but (hyp.) KAB: LCĎ :: MF:NH; and therefore MF having the same ratio to each of the two NH, SR, these are equal (V. 9.) to one another. They are also similar, and similarly situated; therefore GH is equal to PR: and because as AB : CD:: EF : PR, and that PR is equal to GH; therefore AB : CD :: EF: GH; wherefore, if four straight lines, &c.

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PROP. XXIII. THEOR.

EQUIANGULAR parallelograms have to one another the ratio which is compounded of the ratios of their sides.

Let AC, CF be equiangular parallelograms, having the angles BCD, ECG equal : AC has to CF the ratio which is compounded of the ratios of their sides.

Let BC, CG be placed in a straight line; therefore DC, CE are also (1.14.cor.) in a straight line. Complete the parallelogram DG. To DC, CE, CG take (VI. 12.) a fourth proportional K. Then (V. def. 10.) the ratio of BC to K is said to be compounded of the ratios of BC to CG, and of CG to K, or (const.) of the ratios of BC to CG and DC to CE: wherefore BC has to K the ratio compounded of the ratios of the sides. Now (VI. 1.) as

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BC to CG, so is AC to CH; and as DC to CE, so is CH to CF, or (const.) as CG to K, so is CH to CF.' Therefore, since there are two ranks of magnitudes,

BC, CG, K, and AC, CH, CF, which, taken two and two in order, are proportionals, ex æquo, BC is to K, as AC to CF. But it has been shown that BC has to K the ratio compounded of the ratios of the sides ; and therefore AC has to CF the ratio compounded of the ratios of the sides ; that is, of the ratios of BC to CG, and DC to CE. Therefore equiangular parallelograms, &c. *

Cor. Triangles which have one angle of the one equal, or supplemental, to one angle of the other, have to one another the ratio which is compounded of the ratios of the sides containing those angles. For, if BD and EG be joined, the triangles BCD, ECG being (I. 34.) halves of the parallelograms AC, CF, have (V. 15.) the same ratio as the parallelograms : and if BD and CF be joined, the same is true regarding the triangles BCD, CEF, which have the angles BCD, CEF supplemental.

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• Otherwise. - Because there are three parallelograms AC, CH, CF, the first AC (V. def. 10.) has to the third CF, the ratio which is compounded of the ratio of the first AC to the second CH, and of the ratio of CH to the third CF; but AC is to CH, as the straight line BC to CG: and CH is to CF, as the straight line CD is to CE; therefore AC has to CF the ratio which is compounded of ratios that are the same with the ratios of the sides.

The demonstration in the text proceeds on the same principle as that given by Dr. Simson from the Greek of Euclid, but it is considerably simplified by using the lines BC, CG, instead of two lines proportional to them. The foregoing proof, in this note, is given by Simson in his Notes from Candalla. proof, the latter is quite satisfactory; but the other has the advantage of exhibiting in a clear manner the ratio which is compounded of the ratios of the sides, as it is simply that of BC to K.

Dr. Simson remarks in his Note on this proposition, that “nothing is usually reckoned more difficult in the elements of geometry by learners, than the doctrine of compound ratio.” This distinguished geometer, however, has both freed the text of Euclid from the errors introduced by Theon or others, and has explained the subject in such a manner as to remove the difficulties that were formerly felt. According to him, “every proposition in which compound ratio is made use of, may without it be both enunciated and demonstrated ;” and “ the use of compound ratio consists wholly in this, that by means of it, circumlocutions may be avoided, and thereby propositions may be more briefly either enunciated or demonstrated, or both may be done. For instance, if this 23d proposition of the sixth book were to be enunciated, without mentioning compound ratio, it might be done as follows: If two parallelograms be equiangular, and if as a side of the first to a side of the second, so any assumed straight line be made to a second straight line ; and as the other side of the first to the other side of the second, so the second straight line be made to a third : the first parallelogram is to the second, as the first straight line to the third ; and the demonstration would be exactly the same as we now have it. But the ancient geometers, when they observed this enunciation could be made shorter, by giving a name to the ratio which the first straight line has to the last, by which name the intermediate ratios might likewise be signified, of the first to the second, and of the second to the third, and so on, if there were more of them, they called this ratio of

PROP. XXIV. THEOR.

The parallelograms about the diagonal of any parallelogram are similar to the whole, and to one another.

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Let ABCD be a parallelogram, and EG, HK the parallelograms about the diagonal AC: the parallelograms EG, HK are similar to the whole parallelogram, and to one another.

Because DC, G Fare parallels, the angles ADC, AGF are (I. 29.) equal. For the same reason, because BC, EF are parallels, the angles ABC, AEF are equal : and (I. 34.) each of the angles BCD, EFG is equal to the opposite angle DAB, and therefore they are equal to one another: wherefore the parallelograms

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the first to the last, the ratio compounded of the ratios of the first to the second, and of the second to the third straight line : that is, in the present example, of the ratios which are the same with the ratios of the sides."

The proposition in the text will receive illustration from the following one, which exhibits the subject in a different, and in some respects, a preferable light.

Triangles which have an angle of the one equal to an angle of the other, are proportional to the rectangles contained by the sides about those angles : and (2.) equiangular parallelograms are proportional to the rectangles contained by their adjacent sides.

1. Let ABC, DBE be two triangles, having the angles ABC, DBE, equal : the first triangle is to the second as AB.BC is to DB.BE.

Let the triangles be placed with their equal angles coinciding, and join CD. Then (VI. 1.) A B is to DB, as the triangle ABC to DBC. But (VI. 1.) AB : DB : : AB.BC: DB.BC; therefore (V. 11.) AB.BC is to DB. BC, as the triangle ABC to DBC. In the same manner, it would be shown that DB. B C is to DB. BE, as the triangle DBC to DBE; and, therefore, ex æquo, AB. BC is to DB.BE, as the triangle ABC to DBE.

2. If parallels to BC through A and D, and to A B through C and E were drawn, parallelograms would be formed which would be respectively double of the triangles ABC and D BE, and which (V. 15.) would have the same ratio as the triangles ; that is, the ratio of A B.BC to DB.BE : and this proves the second part of the proposition,

Comparing this proposition and the 23d, we see that the ratio which is compounded of the ratios of the sides, is the same as the ratio of their rectangles, or the same (I. 46. cor. 5.) as the ratio of their products, if they be expressed in numbers. This conclusion might also be derived from the proof given in the text. For (const.) DC: CE:: CG: K; whence (VI. 16.) K.DC = CE.CG. But, it was proved that BC: K:: AC : CF; or (VI. 1.) BC. DC : K.DC: : AC: CF; or BC.DC: CE.CG :: AC: CF, because K.DC=CE.CG.

The 14th proposition of this book is evidently a case of this proposition; and the 19th is also easily derived from it.

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