PROPOSITION XXI. THEOREM.

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

Let BAC, EDF, be two triangles.

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

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 angles; and since all the four angles are together equal to four right angles (Book I. Prop. XXVI. Cor. 1.), the remaining two IAH, IDH, will be together equal to two right B

B

H

IE F

H

E

T

G

angles. But the two angles EDF, IDH, are also equal to two right angles: hence the angle EDF is equal to IAй or BAC. In like manner, if the third side EF is perpendicular to the third side BC, it may be shown 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 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.

PROPOSITION XXII. THEOREM.

In any triangle, if a line be drawn parallel to the base, then, all lines drawn from the vertex will divide the base and the allel into proportional parts.

Let DE be parallel to the base BC, and the other lines drawn as in the figure; then will

DI: BF: IK: FG:: KL : GH.

D

A

par

E

I KL

F G HC

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, B 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 same manner we shall find IK FG: KL: GH; and so with the other segments: hence the line DE is divided at the points I, K, L, in the same proportion, as the base BC, at the points F, G, 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 of a right angled triangle, a perpendicu lar be let fall on the hypothenuse; then,

1st. The two partial triangles thus formed, will be similar to each other, and to the whole triangle.

2d. Either side including the right angle will be a mean proportional between the hypothenuse and the adjacent segment. 3d. The perpendicular will be a mean proportional between the two segments of the hypothenuse.

Let BAC be a right angled triangle, and AD perpendicular to the hypothenuse BC.

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 angle C, of the other (Book I. Prop. XXV. Cor 2.): hence those B two triangles are equiangular and

D

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

Secondly. The triangles BAD, BAC, being similar, their homologous sides are proportional. But BD in the small triangle, and BA in the large one, are homologous sides, because they lie opposite the equal angles BAD, BCA; the hypothenuse BA of the small triangle is homologous with the hypothenuse 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 hypothenuse and the segment adjacent to that side.

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

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 AC2 DC.BC; therefore AB2+ AC2=BD.BC+DC.BC= (BD+DC).BC=BC.BC=BC2; or the square described on the hypothenuse BC is equivalent to the squares described on the two sides AB, AC. Thus we again arrive at the property of the square of the hypothenuse, by a path very different from that which formerly conducted us to it and thus it appears that, strictly speaking, the property of the square of the hypothenuse, 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 proposition 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 show that two quantities are equal, we proceed by showing 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

C

of a circle, two chords AB, AC, be drawn to the extremities of a diameter BC, the triangle BAC will be right angled at A (Book III. Prop. B D XVIII. Cor. 2.); hence, first, the perpendicular AD is a mean proportional between the two segments BD, DC, of the diameter, or what is the same, AD2-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 is the same, AB2=BD.BC. In like manner, we have AC2-CD.BC; hence AB2: AC2 :: BD: DC: and comparing AB and AC2, to BC, 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 hypothenuse, have already been given in the third and fourth corollaries of Prop. XI.

PROPOSITION XXIV. THEOREM.

Two triangles having an angle in each equal, are to each other as the rectangles of the sides which contain the equal angles.

In the two triangles ABC, ADE, let the angle A be equal to the angle A; then will the triangle

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

Draw BE. The triangles

ABE, ADE, having the common vertex E, have the same altitude, and consequently are to each other as their bases (Prop. VI. Cor.): that is,

ABE: ADE : : AB : AD. B

In like manner,

D

A

ABC ABE :: AC: AE.

AE

Multiply together the corresponding terms of these proportions, omitting 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 described on their homologous sides.

Let ABC, DEF, be two similar trian- A gles, having the angle A equal to D, and the angle B=E.

Then, first, by reason of the equal an- G gles A and D, according to the last proposition, we shall have

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

B

AB DE AC: DF,

D

H

T

And multiplying the terms of this proportion by the corresponding terms of the identical proportion,

there will result

Consequently,

AC DF: AC: DF,

AB.AC DE.DF: AC2: DF.

ABC DEF :: AC2 : DF2.

Therefore, two similar triangles ABC, DEF, are to each other as the squares described on their 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.