moreover, AC is parallel to DF, the angle ACB will be equal to DFE, and also BAC to EDF; therefore the triangles ABC, DEF, are equiangular and consequently similar.

2. Let the side DE (fig. 124) be perpendicular to AB, and the Fig. 124. side DF to AC. In the quadrilateral AIDH the two angles I, H, will be right angles, and the four angles will be together equal to four right angles (80); therefore the two remaining angles IAH, IDH, are together equal to two right angles. But the two angles EDF, IDH, are together equal to two right angles, consequently the angle EDF is equal to IAH 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 = C, and DEF = B; therefore the two triangles ABC, DEF, which have the sides of the one perpendicular to those of the other, each to each, are equiangular and similar.

210. Scholium. In the first of the above cases the homologous sides are the parallel sides, and in the second the homologous sides are those which are perpendicular to each other. Thus, in the second case, DE is homologous to AB, DF to AC, and EF to BC.

The case of the perpendicular sides admits of the two triangles being differently situated from those represented in figure 124; but the equality of the respective angles may always be proved, either by means of quadrilaterals, such as AIDH, which have two right angles, or by comparing two triangles which, beside the vertical angles, have each a right angle; or we can always suppose, within the triangle ABC, a triangle DEF, the sides of which shall be parallel to those of the triangle to be compared with ABC, and then the demonstration will be the same as that given for the case of figure 124.

THEOREM.

211. Lines AF, AG, &c. (fig. 125), drawn at pleasure through Fig. 125. the vertex of a triangle, divide proportionally the base BC and its parallel DE, so that

DI: BF:: IK: FG:: KL: GH, &c.

Demonstration. Since DI is parallel to BF, the triangles ADI, ABF, are equiangular, and DI: BF :: AI: AF; also, IK being parallel to FG, AI: AF :: IK: FG; hence, on account of

Fig. 126.

the common ratio, AI : AF, DI : BF :: IK : FG. It may be shown, in like manner, that IK: FG:: KL: GH, &c.; therefore the line DE is divided at the points I, K, L, as the base BC is at the points F, G, H.

212. Corollary. If BC should be divided into equal parts at the points F, G, II, the parallel DE would be divided likewise into equal parts at the points I, K, L.

THEOREM.

213. If from the right angle A (fig. 126) of a right-angled tri‐ angle the perpendicular AD be let fall upon the hypothenuse;

1. The two partial triangles ABD, ADC, will be similar to each other and to the whole triangle ABC;

2. Each side AB or AC will be a mean proportional between the hypothenuse BC and the adjacent segment BD or DC;

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

Demonstration. 1. The triangles BAD, BAC, have the angle B common; moreover the right angle_BDA = BAC; consequently the third angle BAD of the one is equal to the third angle C of the other, and the two triangles are equiangular and similar. It may be demonstrated, in the same manner, that the triangle DAC is similar to the triangle BAC; therefore the three triangles are equiangular and similar.

2. Since the triangle BAD is similar to the triangle BAC, their homologous sides are proportional. Now the side BD in the smaller triangle is homologous to the side BA in the larger, because they are opposite to the equal angles, BAD, BCA; the hypothenuse BA of the smaller is homologous to the hypothenuse BC of the larger;

hence

BD: BA:: BA: BC.
In the same manner it may be shown that

DC: AC AC: BC;

therefore each of the sides AB, AC, is a mean proportional between the hypothenuse and the segment adjacent to this side.

3. By comparing the homologous sides of the similar triangles ABD, ADC, we have

BD: AD:: AD: DC;

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

214. Scholium.

The proportion BD:AB::AB: BC, by putting the product of the extremes equal to that of the means, gives

hence

--2

--2

AC=

AB+ AC = BD × BC + DC × BC;

the second member, otherwise expressed, is (BD+DC) × BC,

therefore the square of the hypothenuse BC is equal to the sum of the squares of the two other sides AB, AC. We thus fall' again upon the proposition of the square of the hypothenuse by a process very different from that before pursued; from which it appears, that, properly speaking, the proposition of the square of the hypothenuse is a consequence of the proportionality of the sides of equiangular triangles. Thus the fundamental propositions of geometry reduce themselves, as it were, to this single one, that equiangular triangles have their homologous sides proportional.

It often happens, as in the present instance, that by pursuing the consequences of one or several propositions, we return to the propositions before demonstrated. Generally speaking, that which particularly characterizes the theorems of geometry, and which is an irresistible proof of their certainty, is, that by combining them together in any manner whatever, provided the reasoning be just, we always fall upon accurate results. This would not be the case, if any proposition were false, or only truc to a certain degree; it would often happen, that, by combining the propositions together, the error would augment and become sensible. We have examples of this in all those demonstrations, in which we make use of the reductio ad absurdum. These demonstrations, in which the object is to prove that two quantities are equal, consist in making it evident, that if there were between them the least inequality, we should be led by a course of reasoning to a manifest and palpable absurdity; whence we are obliged to conclude that the two quantities are equal.

Fig. 127.

215. Corollary. If from the point A (fig. 127) of the circumference of a circle two chords AB, AC, be drawn to the extremities of the diameter BC, the triangle ABC will be right-angled at A (128); whence, 1. the perpendioular AD is a mean proportional between the segments BD, DC, of the diameter, or, which amounts to the same thing,

AD = BD × DC.

2. The chord AB is a mean proportional between the diameter BC and the adjacent segment BD;

or,

AB = BD × BC.

-2

2

Also AC DC × BC; therefore AB: AC :: BD: DC. If we

=

These ratios of the squares of the sides to each other and to the square of the hypothenuse have already been given in articles 189, 190.

THEOREM.

216. Two triangles, which have an angle in the one equal to an angle in the other, are to each other as the rectangles of the sides Fig. 128. which contain the equal angles; thus, the triangle ABC (fig. 128) is to the triangle ADE, as the rectangle AB AC is to the rectangle AD × AE.

Demonstration. Draw BE; the two triangles ABE, ADE, whose common vertex is E, have the same altitude, and are to each other as their bases AB, AD (177); hence

In like manner,

ABE: ADE:: AB: AD.

ABC: ABE:: AC: AE;

multiplying the two proportions in order and omitting the common term ABE, we have,

ABC: ADE:: ABX AC: AD × AE.

217. Corollary. The two triangles would be equivalent, if the rectangle AB × AC were equal to the rectangle AD × AE, or if

AB: AD :: AE: AC, which is the case when the line DC is parallel to BE.

THEOREM.

218. Two similar triangles are to each other as the squares of

their homologous sides.

Demonstration. Let the angle A=D (fig. 122), and the an- Fig. 122. gle B= E, then, by the preceding proposition,

ABC: DEF:: ABX AC: DE x DF;

and, because the triangles are similar,

AB: DE:: AC: DF.

This proportion being multiplied in order by the identical proportion.

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.

THEOREM.

219. Two similar polygons are composed of the same number of triangles, which are similar to each other and similarly disposed.

Demonstration. In the polygon ABCDE ( fig. 129) draw from Fig. 129. an angle ♬ the diagonals AC, AD, to the other angles. In the other polygon FGHIK draw, in like manner, from the angle F, homologous to A, the diagonals FH, FI, to the other angles.

Since the polygons are similar, the angle ABC is equal to the homologous angle FGH (162), moreover the sides AB, BC, are proportional to the sides FG, GH, so that

AB: FG:: BC: GH.

It follows from this, that the triangles ABC, FGH, having an angle of the one equal to an angle of the other and the sides about the equal angles proportional, are similar (208), consequently the angle BCA GHF. These equal angles being subtracted from the equal angles BCD, GHI, the remaining