Scholium 1. By the last two propositions, it appears that triangles which are equiangular are similar: and conversely: if triangles have their sides proportional, they are equiangular, and consequently, similar.

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 changing the angles, or the angles may be changed without altering the proportion between the sides. Thus, in quadrilaterals, equality between the corresponding angles does not insure proportionality among the sides: and reciprocally: proportionality among the sides does not insure equality among the corresponding angles. It is evident, for example, that if in the quadrilateral ABCD, we draw EF parallel to BC, the angles of the quadrilateral AEFD, are made equal to those of ABCD; though the proportion between their sides is different; and in like manner, without changing the four sides AB, BC, CD, AD, we can change the angles by making the point B approach to D, or recede from it.

D

F

C

A

E B

Scholium 2. The two preceding propositions, are in strictness but one, and these, 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 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.

Let ABC, DEF, be two triangles, having the angle A equal to D; then, if

the two triangles will be similar.

Make AG=DE, and draw GH A parallel to BC. The angle AGH will be equal to the angle ABC (B. I., P. 20, c. 3); and the triangles AGH, ABC, will be equiangular: hence, we shall have,

B

AB : AG : AC : AH.

But, by hypothesis, we have,

AB : DE: AC: DF;

and by construction, AG-DE: hence AH-DF. Therefore, the two triangles AGH, DEF, have two sides and the included angle of the one equal to the sides and the included angle of the other: hence, they are equal (B. I., P. 5); but the triangle AGH is similar to ABC: therefore, DEF is also similar to ABC.

PROPOSITION XXI. THEOREM.

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

Let BAC, EDF, be two triangles, having their sides respectively parallel to each other; then will they be similar.

First. If the side BA is parallel to ED, and BC to EF, the angle ABC is equal to DEF (B. I., P. 24): if CA is parallel to FD, the angle BCA is equal to EFD, and also, BAC to EDF; hence, the triangles CBA, FED, are equiangular; consequently they are similar (P. 18).

G

A

D

H

IE F

B

C

ing two IAH, IDH, are together equal to two right angles. But the two angles EDF, IDH, are also equal to two right angles (B. I., P. 1): hence, the angle EDF is equal to IAH, or BAC (A. 3). 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 (P. 18).

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 BA, DF with AC, and EF with BC.

The case of the perpendicular sides may present a relative position of the two triangles different from that exhibited in the diagram. But we can always conceive a triangle FED to be constructed within the triangle ABC, and such that its sides shall be parallel to those of the triangle compared with BAC; and then the demonstration given in the text will apply.

PROPOSITION XXII. THEOREM.

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

Let BAC be a triangle, DE parallel to the base BC, and the other lines drawn as in the figure;

then will

KL : GH.

hence (B. II., P. 4, c.), DI : BF :: IK: FG.

In the same manner, we may prove that

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 is divided, 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 be divided also into equal parts at the points I, K, L.

PROPOSITION XXIII. THEOREM.

In a right-angled triangle, if a perpendicular is drawn from the vertex of the right angle to the hypothenuse.

1st. The triangles on each side of the perpendicular are similar to the whole triangle, and to each other:

2d. Either side about the right angle is a mean proportional between the hypothenuse and the adjacent segment:

3d. The perpendicular is a mean proportional between the segments of the hypothenuse.

Let BAC be a right-angled triangle, and AD perpen

dicular 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 (B. I., P. 25, c. 2):

B

A

D

hence, these two triangles are similar (P. 18). In the same

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

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

B

A

D

gous 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 (P. 18, s.); 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 have

DC : AC AC: BC;

::

hence, each of the sides AB, AC, is a mean proportional between the hypothenuse and the adjacent segment.

Thirdly. Since the triangles DBA, DAC, are similar, we have, by comparing their homologous sides,

BD AD :: AD: DC;

:

hence, the perpendicular AD is a mean proportional between the segments BD, DC, of the hypothenuse.

Scholium. Since BD : AB :: AB : BC,

we have (B. II., P. 1, c.), AB

BDX BC.

For the same reason, AC2 DCX BC;

therefore, AB2+AC2≈≈BD×BC+DC×BC÷~(BD+DC)×

BC=BC×BC÷BC2 ;

that is, the square described on the hypothenuse BC is equiva lent to the sum of the squares described on the two sides BA, AC. Thus, we again arrive at this property of the right-angled triangle, and by a path very different from that which formerly conducted us to it: and thus it appears that, strictly speaking, this property 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 equi angular triangles have their homologous sides proportional.