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Two equiangular triangles have their homologous sides proportional, and are similar.

Let ABC, CDE be two triangles which have their angles equal each to each, namely, BAC-CDE, ABC-DCE and ACB-DEC; then the homologous sides, or the sides adjacent to the equal angles, will be proportional, so that we shall have BC: CE :: AB : CD :: AC : DE.


Place the homologous sides BC, CE in the same straight line; and produce the sides BA, ED, till they meet in F.

Since BCE is a straight line, and the angle BCA is equal to CED, it follows that AC is parallel to DE (Book I. Prop. XIX. Cor. 2.). In like manner, since the angle ABC is equal to DCE, the line AB is parallel to DC. Hence the figure ACDF is a parallelogram.

In the triangle BFE, the line AC is parallel to the base FE; hence we have BC: CE: BA: AF (Prop. XV.); or putting CD in the place of its equal AF,


In the same triangle BEF, CD is parallel to BF which may be considered as the base; and we have the proportion BC: CE:: FD: DE; or putting AC in the place of its equal FD, BC CE: AC: DE.

And finally, since both these proportions contain the same ratio BC ČE, we have

AC: DE: BA : CD.

Thus the equiangular triangles BAC, CED, have their homologous sides proportional. But two figures are similar when they have their angles equal, each to each, and their homologous sides proportional (Def. 1.); consequently the equiangular triangles BAC, CED, are two similar figures.

Cor. For the similarity of two triangles, it is enough that they have two angles equal, each to each; since then, the third will also be equal, in both, and the two triangles will be equiangular.

Scholium. Observe, that in similar triangles, the homologous sides are opposite to the equal angles; thus the angle ACB being equal to DEC, the side AB is homologous to DC; in like manner, AC and DE are homologous, because opposite to the equal angles ABC, DCE. When the homologous sides are determined, it is easy to form the proportions:

AB: DC: AC: DE:: BC: CE.


Two triangles, which have their homologous sides proportional, are equiangular and similar.

In the two triangles BAC, DEF, suppose we have BC: EF :: AB : DE :: AC: DF; then will the triangles ABC, DEF have their angles_equal, namely, A=D, B=E, C=F.

At the point E, make the angle




FEG B, and at F, the angle EFG-C; the third G will be equal to the third A, and the two triangles ABC, EFG will be equiangular (Book I. Prop. XXV. Cor. 2.). Therefore, by the last theorem, we shall have BC: EF : :AB: EG; but, by hypothesis, we have BC: EF:: AB: DE; hence EG-DE. By the same theorem, we shall also have BC: EF:: AC : FG; and by hypothesis, we have BC EF:: AC: DF; hence FG-DF. Hence the triangles EGF, DEF, having their three sides equal, each to each, are themselves equal (Book I. Prop. X.). But by construction, the triangles EGF and ABC are equiangular : hence DEF and ABC are also equiangular and similar.

Scholium 1. By the last two propositions, it appears that in triangles, equality among the angles is a consequence of proportionality among the sides, and conversely; so that either of those conditions sufficiently determines the similarity of two triangles. The case is different with regard to figures of more than three sides: even in quadrilaterals, the proportion between the sides may be altered without altering the angles, or the angles may be altered without altering the proportion between the sides; and thus proportionality among the sides cannot be a consequence of equality among the angles of two quadrilaterals, or vice versa. It is evident, for example, that H

by drawing EF parallel to BC, the angles of the quadrilateral AEFD, are made equal to those of ABCD, though the proportion between the sides is different; and, in like manner, without changing the four sides AB, BC, CD, AD, we can make the point B approach D or recede from it, which will change the angles.



Scholium 2. The two preceding propositions, which in strictness form but one, together with that relating to the square of the hypothenuse, are the most important and fertile in results of any in geometry: they are almost sufficient of themselves for every application to subsequent reasoning, and for solving every problem. The reason is, that all figures may be divided into triangles, and any triangle into two right angled triangles. Thus the general properties of triangles include, by implication, those of all figures.


Two triangles, which have an angle of the one equal to an angle of the other, and the sides containing those angles proportional, are similar.

In the two triangles ABC, DEF, let the angles A and D be equal; then, if AB : DE :: AC: DF, the two triangles will be similar.

Take AG-DE, and draw GH parallel to BC. The angle AGH will be equal to the angle ABC (Book I. Prop. XX.




Cor 3.); and the triangles AGH, ABC, will be equiangular : hence we shall have AB: AG: AC: AH. But by hypothesis, we have AB DE AC: DF; and by construction, AG=DE: hence AH-DF. The two triangles AGH, DEF, have an equal angle included between equal sides; therefore they are equal; but the triangle AGH is similar to ABC; therefore DEF is also similar to ABC.


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 BAC 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




angles. But the two angles EDF, IDH, are also equal to two right angles: hence the angle EDF is equal to IAĤ 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 homològous 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.


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 parallel 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.

For, since DI is parallel to BF, the triangles ADI and ABF are equiangular; and we have DI: BF:: ΑΙ AF; and since IK is parallel to FG, B we have in like manner AL: 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 Ï, K, L.


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 two triangles are equiangular and


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