PB is to pb as A B to ab, or as B C to bc; that is, alternando, PB: BC:: pb: bc. And, in the same manner, if any other corresponding sides, as EF and ef be divided in the same ratio in the points Q, q, it may be shown that QE: ED qe.ed, or, invertendo, that DE:EQ::de: eq. Therefore, by Cor. 1., the straight lines PQ and pq, which join the points P, Q and p, q, make equal angles with A B, a b respectively, and with E F, ef, respectively, and PQ is to pq as PB to pb, or as AB to a b.

PROP. 33. (EUc. vi. 7.)

If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals; and, if one of the remaining angles be a right angle, or if they be both greater, or both less than right angles; the two triangles shall be similar.*

Let ABC, DEF be two triangles, which have the angle at A equal to the angle at D, and the sides about two other angles, B and E, proportionals: then, if one of the remaining angles, as C, be a right

44

angle, or if both of them, C, F, be greater, or both less, than right angles, the triangle ABC shall be similar to the triangle DEF.

At the point E make the angle DEG equal to AB C, and let the line E G meet DF, or D F produced in G. Then, because the triangles A B C, DEG have two angles of the one equal to two angles of the other, each to each, they are equiangular: therefore (31.) DE: EG ::AB: BC, but AB: BC::DE:EF: therefore (12.) DE:EG::DE:EF, and (18.) E G is equal to E F.

And first, if one of the angles ACB, DFE be a right angle, whether it be ACB or DFE, one of the lines EG, EF will be at right angles to D F; and therefore, if EG do not coincide with EF, there will be an isosceles triangle EFG, which has one of the angles at the base a right angle, and therefore (I. 6.) the other likewise a right angle, and the two together equal to two right

Angles which are both greater or both less than right angles are said to be of the same affection: hence, instead of saying "both greater or both less than right angles," this proposition is sometimes enunciated by saying "both of the same affection."

angles, which is impossible (I. 8.): therefore, in this case, EG cannot but coincide with E F, that is, the angle DEF cannot but be equal to the angle DEG, that is, to AB Č.

Again, if the angles A CB, DFE be both greater than right angles, they will be together greater than two right angles; or, if they be both less than right angles, they will be together less than two right angles: but, if EG do not coincide with EF, these same two angles will be together equal to the angles D G E, DFE, that is, to the angles DGE, EGF, (I. 6.) because EG is equal to EF, that is, to two right angles (I. 2.),-which is impossible: therefore in these cases, also, EG cannot but coincide with E F, that is, the angle D E F cannot but be equal to the angle DEG, that is, to AB C.

And, because the triangles ABC, DEF have two angles of the one equal to two angles of the other, each to each, they are equiangular (I. 19. Cor. 1.), and therefore similar (31. Cor. 1.). Therefore, &c.

PROP. 34. (EUC. vi. 8.)

In a right-angled triangle, if a perpendicular be drawn from the right angle to the hypotenuse, the triangles upon either side of it shall be similar to the whole triangle, and to one another.

Let A B C be a right-angled triangle, and from the right angle A, let there be drawn to the hypotenuse the perpendicular AD: the triangles DBA, DAC shall be similar to the whole triangle ABC, and to one another.

Because the angles B D A, B A C are right angles, and because the angle at B is common to the two triangles DBA, ABC, their third angles (I. 19. Cor. 1.) are equal to one another; and therefore the triangle DBA, being equiangular with the triangle ABC, is similar to it (31.Cor. 1.). In the same manner it may be shown that DAC is equiangular with A B C, and therefore similar to it. And, because DBA, DAC are equiangular with the same triangle, they are equiangular with, and therefore similar to, one another.

Therefore, &c.

Cor. (Euc. vi. 8. Cor.) The perpendicular AD is a mean proportional between the segments BD, DC of the hypotenuse; and either side, as A B, is a mean proportional between the

whole hypotenuse B C, and the segment BĎ, which is adjacent to it.

For, because the angles BAD, ACD, are equal to one another, the sides BD, AD of the similar triangles BDA, ADC are homologous, and the angles BDA, ADC are equal to one another; therefore (31.) BD: DA :: AD or DÁ: DC: and, in like manner, because the sides BD, BA of the similar triangles BDA, BAC are homologous, and the angles at B equal, BD:BA::BA:B C.

SECTION 5.-Proportion of the Surfaces of Reetilineal Figures.

PROP. 35.

Rectangles which have the same or equal altitudes, are to one another as their bases.

For, if the base of one of the rectangles be divided into any number of equal parts, the rectangle itself will be divided into as many equal rectangles by straight lines drawn parallel to its side through the points of division. Also, the base of the other rectangle will contain a certain number of parts equal to those into which the first base is divided, exactly or with an excess less than one of those parts, and that rectangle will contain as many rectangles equal to those into which the first rectangle is divided, exactly or with a corresponding excess less than one of them. And this will be the case whatsoever be the number of parts in the first base and rectangle. Therefore, (def. 7.) the first rectangle is to the second as the base of the first is to the base of the second; that is, rectangles of the same altitude are to one another as their bases.

Therefore, &c.

The same demonstration has been already given, with a figure, to illustrate the definition of Proportionals in Section 3. (See def. 7.)

Cor. (Euc. vi. 1 part of.) In the same manner it may be shown, that any two parallelograms which have the same or equal altitudes are to one another as their bases; or, the same may be directly inferred from this proposition, for every parallelogram is equal to a rectangle, having the same base and altitude.

Scholium.

Proportions of this kind, in which the two first terms stand in the same relation to the two last respectively, as, in the above instance, in the relation of a

rectangle of given altitude to its base, are of perpetual occurrence in the Mixed Sciences, and are commonly enunciated, as in the proposition, by the word as. They constitute, indeed, a peculiar class, which has received the name of Variations; one quantity being said to vary as another when it increases and decreases in the same proportion with that other. We have already had one instance in a rectangle of given altitude and its base. To take another perhaps more familiar, we say commonly, that the weight of a mass of lead is in proportion to its magnitude; and the same is understood, when it is said, that the weight of such a mass is as its magnitude, or varies as its magnitude. Again, the distance a horse runs in a given time "is in proportion to his speed," or "is as his speed," or "varies as his speed." It is evident that the use of the word as in these and the like cases enables us to avoid the following and similar enunciations. "The weight of one mass of lead is to the weight of another, as the magnitude of the first to the magnitude of that other." "The distance a horse runs in a given time with one velocity, is to the distance he would run in the same time with another velocity, as the first velocity to that other." (See Arithmetic, art. 136.)

PROP. 36.

Any two rectangles are to one another in the ratio which is compounded of the ratios of their sides.

A

Let A C, E F be two rectangles, and let an angle of the one be made to coincide (I. 1. and ax. 11.) with an angle of the other, as at B: the rectangle AC shall be to the rectangle EF in the ratio which is compounded of the ratios of AB to EB, and of B C to B F.

E

B

For the rectangle A C is to the rectangle E F in the ratio which is compounded of the ratios of AC to EC, and of EC to E F. (def. 12.) But, because the rectangles A C, EC have the same altitude BC, AC is to EC as A B to E B (35.); and, in like manner, because the rectangles EC, EF have the same altitude E B, E C is to EF as BC to BF. Therefore, (27. Cor. 1.) the ratio which is compounded of the ratios of A C to E C, and of EC to EF, is the same with the ratio, which is com

pounded of the ratios of AB to EB, and BC to B F; that is, the rectangle AC is to the rectangle E F in the ratio which is compounded of the ratios of AB to E B, and of B C to EF.

Therefore, &c.

Cor. 1. (Euc. vi. 23.) In the same manner it may be shown that any two parallelograms AC, EF, which have one angle of the one equal to one angle

of the other, are to one another in the ratio which is compounded of the ratios of the sides about the equal angles. For if the equal angles be made to

coincide, as at B, the parallelogram AC is to the parallelogram E F in the ratio which is compounded of the ratios of AC to EC, and of EC to EF, that is (35. Cor.) in the ratio which is compounded of the ratios of A B to EB, and of B C to BF.

Cor. 2. It appears, also, from the proposition, that any two parallelograms whatever are to one another in the ratio which is compounded of the ratio of their bases and altitudes; for the rectangles, to which (I. 24. Cor.) they are equal, are in that ratio.

PROP. 37

If the straight line A be to the straight line B in one ratio, and the straight line A' to the straight line B' in another ratio, the rectangle A A' shall be to the rectangle B B in the ratio which is compounded of these two ratios. This proposition is the same with the preceding, which is here stated under a different form, as well for the sake of rendering more obvious such references as may be made to it under this form, as to separate the following corollaries from those of Prop. 36. Its demonstration is accordingly contained in that of Prop. 36. Cor. 1. If A be to B as A' to B', the rectangle A A' will be to the rectangle B B' in the duplicate ratio of A to B. For, if C be taken a third proportional to A and B, then, because B is to C as A to B, that is as A' to B', the ratio which is compounded of the ratios of A to B and of A' to B' is the same with the ratio of A to C, that is (Def. 11.) with the duplicate ratio of A to B.

Cor. 2. Hence if three straight lines A, B, C, be proportionals, the square of the first shall be to the square of the second as the first is to the third; that is, the duplicate ratio of two straight lines is the same with the ratio of their squares,

Cor. 3. If A, B, C, D be four straight lines, and A' B', C', D' four others, and if ABCD

and A': B':: C': D'

then AA': BB' :: CC': DD'; that is, if four straight lines A, B, C, D be proportionals, and likewise four others A', B', C', D'; the rectangles A A', B B',

C C', D D', which are contained under

the corresponding antecedents and consequents, shall also be proportionals. For ratios which are compounded of the same ratios are the same with one another (27). And hence it is evident, conversely, that if AA', BB', CC', DD' be proportionals, and A', B', C', D' likewise proportionals; A, B, C, D must also be proportionals.

Cor. 4. If four straight lines A, B, C, D be proportionals, their squares A, B2, C2, Do shall likewise be proportionals; and conversely.

PROP. 38. (EUc. vi. 16.)

If four straight lines be proportionals the rectangle contained by the extremes shall be equal to the rectangle contained by the means; and conversely, if the rectangle contained by two straight lines be equal to the rectangle contained by other two, the four straight lines shall constitute a proportion, in which the sides of one rectangle are extremes, and the sides of the other rectangle means.

The first part of this proposition has been already demonstrated the Scholium on Prop. 28., Sect. 3.; which demonstration the reader is here desired to consult, as depending immediately upon the definition of proportionals in Sect. 3.

So important a theorem, however, cannot be considered in too many points The following proof, by help of view. of Prop. 35., is therefore added, being in effect the same with that of Euclid.

B

Let AB, AC, AD, AE be the four proportionals, of which the extremes AB, AE are placed at right angles to one another, and contain the rectangle B E, and the means AC, AD are placed in BA, EA produced, and, being for that reason (1.3.) likewise at right angles to one another, contain the rectangle CD. Complete the rectangle CE. Then, because the rectangles BE, CE have the same

altitude AE, the rectangle BE is to the rectangle CE as AB to AC (35.): and, in like manner, because the rectangles CD, CE have the same altitude AC, the rect

angle CD is to the rectangle CE as AD to A E; but the line AB is to AC as AD to AE; therefore, the rectangle BE is to CE as CD to CE. And, because the rectangles BE, CD have the same ratio to the same rectangle CE, they are equal to one another (11. Cor. 1.). Next, let the rectangle under AB, AE, be equal to the rectangle under AC, AD; the lines AB, AC, AD, AE, shall be proportionals, AB, AE being extremes, and AC, AD means.

For, the rectangles being placed as before, and the rectangle CE being completed, BE is to CE as CD to CE (10.); but BE is to CE as AB to AC (35.), and CD to CE as AD to AE; therefore (12.) AB is to AC as AD to AE.*

Cor. 2. It appears from the proposition that rectangles which have their sides about the right angles reciprocally proportional, are equal; and, conversely, that equal rectangles have their sides about the right angles reciprocally proportional.

Cor. 3. (Euc. vi. 14.) And, in the same manner, it may be shown, that any two parallelograms which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another; and conversely. For, if the equal angles be placed vertically as in the figure, and the parallelogram CE be completed, the demonstration of this more general case will be the same with that

D

Or we may say, “AD: AE::AB: AC," in which case AC and AD will be extremes, and A B, AE means.

of the proposition; 35. Cor. being cited instead of 35.

Scholium.

By help of this proposition the proportion of four lines is convertible into the equality of two rectangles, and the reverse. Thus it appears, that I. 36. Cor. 2. and 3., in which it is inferred that the squares of the sides of a right angled triangle are respectively equal to the rectangles under the hypotenuse and its adjoining segments, and that the the rectangle under the segments of the square of the perpendicular is equal to hypotenuse, may be stated in the words

of 34. Cor., in which it is inferred that either side is a mean proportional between the hypotenuse and segment adjacent to it, and that the perpendicular is a mean proportional between the segments of the hypotenuse. Again, I. 38. in which it is demonstrated that, in every triangle, if a perpendicular be drawn from the vertex to the base, or to the base

produced, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base, or of the base produced, may be stated thus:-the base is to the sum of the sides ence of the segments of the base, or sum of as the difference of the sides to the differthe segments of the base produced. For it is shown (I. 38.) that the difference of the squares of two straight lines is equal to the rectangle under their sum and dif

ference.

Many other instances will occur in the remaining part of this treatise, in which the demonstrations are considerably abridged by the use of this very important theorem.

We shall conclude the present Scholium by applying it to the demonstration of the following Lemmas (or auxiliary theorems) which will be found of service in such problems as have the sum or the difference of the sides of a triangle among their data. (See Book III. Sect. 7.) It will be seen that they belong, according to our arrangement, to the subject of the preceding Section; to which they should have been subjoined, had they admitted of an easy demonstration without the aid of this 38th Proposition.

Lemma 1.

If a perpendicular be drawn from the vertex of a triangle to the base, and if the base be equally produced both ways, so that the base produced may be a third proportional to the base and the

Bisect BC in G: then because EB is equal to CF, E F is likewise bisected in G. Therefore the difference of ED, DF is equal to twice GD. But twice GD is equal to the difference or to the sum of BD, DC, according as the point D is in BC, or in BC produced. Therefore the difference of ED, D F, is equal to the difference of BD, DC, or to the sum of BD, DC, according as the point D lies in BC, or in BC produced. Again, because AD is perpendicular to BC, the difference of the squares of BA, AC is equal to the difference of the squares of BD, DC (I. 38.) or, which is the same thing (1.34.), the rectangle under the sum and difference of BA, AC is equal to the rectangle under the sum and difference of BD, DC. Therefore (38.) BA~AC: BDDC:: BD+DC: BA+AC;* that is, since the second term is equal to ED-DF, and the third to BC, BAAC: ED-DF :: BC: BA + AC, that is,:: BA+AC: ED+DF, by the supposition. Therefore, alternando (19.) BA-AC: BA+AC::ED-DF: ED +DF: and, by sum and difference (24. Cor. 2.) 2 BA: 2AC::2ED: 2 DF, that is, (17.Cor. 2.) BA: AC:: ED: DF. Therefore, &c.

Cor. If DE be taken to AB as BA+ AC to BC, or, which is the same thing, in a ratio which is the subduplicate of GE to GB, then the other segment DF of the base equally produced in the opposite direction shall be to the other side

AC in the same ratio.

The sign deuotes that the sum or difference is to be taken according to one or other of two supposed cases, that is, in the present instance, according as the point D lies in B C, or in B C produced; the sign that the difference or sum is to be taken according to those cases.

Lemma 2.

If a perpendicular be drawn from the the base be equally reduced both ways, vertex of a triangle to the base, and if so that the base reduced may be a third proportional to the base and the differ

ence of the sides, the sides shall be to one another as the corresponding segments of the base reduced.

The demonstration is so similar to that of Lemma 1., that it will be readily apprehended from the following outline, with reference to the adjoined figures.

ABC is the triangle, AD the perpendicular upon the base, EF the base

equally reduced both ways, so that EF BA-AC:BC, and G is the middle point of the base.

Then, because BE=

CF, the point G bisects also E F, and ED+DF= 2 GD, that is, = BD DC, according as the point D lies in BC, or in AC-BD-DC into a proportion, as in BC produced. Now, converting BA Lem. 1, BA + AC: BD DC :: BD± DC: BA-AC, or BA+ACED+ DF:: BC: BA-AC, that is, :: BA

~AC: EF (or EDDF). Therefore

alternando, BA+AC: BA~AC:: ED +DF: ED-DF: and hence, by sum and difference, BA: AC:: ED: DF. Therefore, &c.

PROP. 39. (EUc. vi. 1., first part of.)

Triangles which have the same or equal altitudes are to one another as their

bases.

For such triangles are the halves of rectangles which are upon the same bases respectively, and have the same or equal altitudes: and because these rectangles are as the bases (35.), the triangles, which are their halves, have to one another the same ratio (17. Cor. 1.).

Otherwise: As in 35. it is demonstrated of the rectangles, so here it may, after the same manner, be deof the monstrated two triangles A B C, AD E, and the two bases B C, D E, that they are propor

C D

F