equal sides; therefore they are equal; but the triangle AGH is similar to ABC; therefore DEF is also similar to ABC.

PROPOSITION XXI. THEOREM.

Two triangles, which have their homologous sides parallel, or perpendicular to each other, are similar.

First. If the side AB is parallel to DE, and BC to EF, the angle ABC (26. I.) will be equal to DEF; if AC is parallel to DF, the angle ACB will be equal to DFE, and also BAC to EDF; hence the triangles ABC, DEF are equiangular; hence they are similar.

B

H

G

E

Secondly. If the side DE is perpendicular to AB, and the side DF to AC, the two angles I and H of the quadrilateral AIDH will be right; and since (28. I.) all, the four angles are together equal to four right angles, the remaining two IAH, IDH will be together equal to two right angles. But the two angles EDF, IDH are also equal to two right angles: hence the angle EDF is equal to IAH or BAC. In like manner, if the third side EF is perpendicular to the third BC, it may be shewn that the angle DFE is equal to C, and DEF to B: hence the triangles ABC, DEF, which have the sides of the one perpendicular to the corresponding sides of the other, are equiangular and similar.

Scholium. In the case of the sides being parallel, the homologous sides are the parallel ones: in the case of their being perpendicular, the homologous sides are the perpendicular ones. Thus in the latter case DE is homologous with AB, DF with AC, and EF with BC.

The case of the perpendicular sides might present a relative position of the two triangles different from that exhibited in the diagram: but the equality of the respective angles might still be demonstrated, either by means of quadrilaterals like AIDH having two right angles, or by the comparison of two triangles having two of their angles vertical, and a right angle in each. Besides, we might always conceive a triangle DEF to be constructed within the triangle ABC, and such that its sides should be parallel to those of the triangle compared with ABC; and then the demonstration given in the text would apply.

E

PROPOSITION XXII. THEOREM.

The lines AF, AG, &c. drawn anyhow through the vertex of a triangle, will divide BC the base, and DE parallel to the base, in the same proportion; so that we shall have DI: BF:: IK: FG:: KL: GH, &c.

D

For since DI is parallel to BF, the triangles ADI and ABF are equiangular; and we have DI: BF:: AI: AF; and since IK is parallel to FG, we have in like manner AI: AF::IK: FG; hence, the ratio AI: AF being common, we shall have DI: BF:: IK: FG. In the shall find B IK: FG:: KL: GH; and so with the other segments: hence the line DE is divided at the points I, K, L, as the base BC at the points F, G, H.

same manner

we

H

Cor. Therefore if BC were divided into equal parts at the points F, G, H, the parallel DE would also be divided into equal parts at the points I, K, L.

PROPOSITION XXIII. THEOREM.

If from the right angle A of a right-angled triangle, the perpendicular AD be let fall on the hypotenuse; then, First, The two partial triangles ABD, ADC, will be similar to each other and to the whole triangle ABC;

Secondly, Either side AB or AC will be a mean proportional between the hypotenuse BC and the adjacent segment BD or DC; and,

Thirdly, The perpendicular AD will be a mean proportional between the two segments BD and DC.

First. The triangles BAD and BAC have the common angle B, the right angle BDA BAC, and therefore the third angle BAD of the one equal to the

third C of the other: hence those two tri- B D

angles are equiangular and similar. In the

same manner it may be shewn that the triangles DAC and BAC are similar; hence all the three triangles are similar and equiangular.

Secondly. The triangles BAD, BAC being similar, their ho

mologous sides are proportional. But BD in the little triangle and BA in the large one are homologous, because they lie opposite the equal angles BAD, BCA; the hypotenuse BA of the little triangle is homologous with the hypotenuse BC of the large triangle: hence the proportion BD: BA :: BA : BC. By the same reasoning, we should find DC: AC :: AC: BC; hence each of the sides AB, AC is a mean proportional between the hypotenuse and the segment adjacent to that side.

Thirdly. Since the triangles ABD, ADC are similar, by comparing their homologous sides, we have BD: AD: : AD: DC; hence, the perpendicular AD is a mean proportional between the segments DB, DC of the hypotenuse.

Scholium. Since BD: AB::AB: BC, the product of the extremes will be equal to that of the means, or AB2=BD.BC. For the same reason we have AC-DC.BC; therefore AB2 +AC2=BD.BC + DC.BC=(BD+DC).BC=BC.BC=BC2; or the square described on the hypotenuse BC is equal to the squares described on the two sides AB, AC. Thus we again arrive at the property of the square of the hypotenuse, by a path very different from that which formerly conducted us to it; and thus it appears that, in strict speaking, the property of the square of the hypotenuse is a consequence of the more general property, that the sides of equiangular triangles are proportional. Thus the fundamental propositions of geometry are reduced, as it were, to this single one, that equiangular triangles have their homologous sides proportional.

It happens frequently, as in this instance, that by deducing consequences from one or more propositions, we are led back to some proposition already proved. In fact, the chief characteristic of geometrical theorems, and one indubitable proof of their certainty is, that, however we combine them together, provided only our reasoning be correct, the results we obtain are always perfectly accurate. The case would be different, if any propo

sition were false or only approximately true; it would frequently happen that on combining the propositions together, the error would increase and become perceptible. Examples of this are to be seen in all the demonstrations, in which the reductio ad absurdum is employed. In such demonstrations, where the object is to shew that two quantities are equal, we proceed by shewing that if there existed the smallest inequality between the quantities, a train of accurate reasoning would lead us to a manifest and palpable absurdity; from which we are forced to conclude that the two quantities are equal.

Cor. If from a point A, in the circumference of a circle, two chords AB, AC be drawn to the extremities of a diameter BC, the triangle BAC

(Prop. 18. II.) will be right-angled at A; hence, BD

first, the perpendicular AD is a mean proportional between the two segments BD, DC, of the diameter, or what amounts to the same, AD-BD.DC.

Hence also, in the second place, the chord AB is a mean proportional between the diameter BC and the adjacent segment BD, or what amounts to the same, AB2=BD.BČ. In like manner, we have AC2=CD.BC; hence AB2: AC2:: BD: DC; and comparing AB2 and AC2 to BC2, we have AB2: BC2:: BD: BC, and AC2: BC2:: DC: BC. Those proportions between the squares of the sides compared with each other, or with the square of the hypotenuse, have already been given in the third and fourth corollaries of Prop. 11.

PROPOSITION XXIV. THEOREM.

Two triangles, having an equal angle, are to each other as the rectangles of the sides which contain that angle. Thus the triangle ABC is to the triangle ADE as the rectangle AB.AC is to the rectangle AD.AE.

Draw BE. The triangles ABE, ADE, having the common vertex E, have the same altitude, and consequently (Prop. 6.) are to each other as their bases: that is,

ABE: ADE:: AB: AD.

In like manner,

ABC ABE:: AC: AE.

A

D

B

E

B

Multiply together the corresponding terms of those proportions, omit ting the common term ABE; we have

ABC: ADE :: AB.AC: AD.AE.

Cor. Hence the two triangles would be equivalent, if the rectangle AB.AC were equal to the rectangle AD.AE, or if we had AB: AD:: AE: AC; which would happen if DC were parallel to BE.

PROPOSITION XXV. THEOREM.

Two similar triangles are to each other as the squares of their homologous sides.

Let the angle A be equal to D, and the angle B-E. Then, first, by reason of the equal angles A and D, according to the last proposition, we shall have

ABC: DEF:: AB.AC: DE.DF.

Also, because the triangles are similar,
AB: DE: AC: DF.

D

And multiplying the terms of this proportion by the corre sponding terms of the identical proportion,

AC: DF:: AC: DF,

Therefore two similar triangles ABC, DEF are to each other as the squares of the homologous sides AC, DF, or as the squares of any other two homologous sides.

PROPOSITION XXVI. THEOREM.

Two similar polygons are composed of the same number of triangles similar each to each, and similarly situated.

These polygons being similar, the angles ABC, FGH, which are homologous, must be equal (Def. 2.), and the sides AB, BC must also be proportional to FG, GH, that is, AB: FG:: BC: GH. Wherefore the triangles ABC, FGH have each an equal angle, contained between proportional sides; hence they are similar; hence the angle BCA is equal to GHF. Take away these equal angles from the equal angles BCD, GHI; there remains ACD=FHI. But since the triangles ABC, FGH are similar, we have AC: FH:: BC: GH; and (Def. 2.) since the polygons are similar, BC: GH:: CD: HI; hence AC:FH::CD:HI. But the angle ACD, we already know, is equal to FHI; hence