Euclid's Elements of geometry, the first three books (the fourth, fifth, and sixth books) tr. from the Lat. To which is added, A compendium of algebra (A compendium of trigonometry).

COR. 1.-If in a triangle ABC a line DE be drawn parallel to any side AC, the triangle cut off, DBE, will be similar to the whole.

For on account of the common angle B, and the angles BDE, BED being equal to the internal angles at the same side, RAC, BCA (by Prop. 29, B. 1), it is equiangular, and therefore similar (by Schol.).

COR. 2.-If there be drawn a parallel DE to any side AC of the triangle ABC, the right line BO drawn from the opposite angle shall cut the parallels into proportional parts.

For since in the triangle ABO, DI is parallel to AO, AO will be to DI as OB to IB (by the Schol. of this Prop.); and similarly, OC is to IE as OB to IB; and therefore, AO is to DI as OC to IE (by Prop. 18, B. 5), and by alternation, AO is to OC as DI to IE.

If two triangles (ABC, DEF) have the sides proportional (BA to AC as ED to DF, and AC to CB as DF to FE), they will be equiangular; and the equal angles are subtended by homologous sides.

At the extremities of any side DE of either triangle DEF let the angles EDG and DEG be constructed, equal to the angles A and B, at the extremities of the side AB, which is homologous to ED; and the remaining angle

G in the triangle DEG will be equal to the remaining angle C in the triangle ABC (by Cor. 2, Prop. 32, B. 1). Therefore, since the triangle ABC is equiangular to the triangle DEG (by Const.), BA will be to AC as ED to DG (by Prop. 4, B. 6); but BA is to AC as ED to DF (by Hypoth.), and therefore ED is to DG as ED is to DF (by Prop. 18, B. 5), therefore DG and DF are equal (by Prop. 15, B. 5); it can be similarly demonstrated that EG and EF are equal, therefore the triangle EDG is equilateral to the triangle EDF, and therefore equiangular (by Schol. Prop. 8, B. 1); but EDG is equiangular to BAC (by Const.), and therefore BAC is also equiangular to EDF; it is evident that the equal angles are subtended by the homologous sides.

If two triangles (ABC and DEF) have one angle of the one equal to one angle of the other, (A to D) and the sides about the equal angles proportional (BA to AC as ED to DF), the triangles will be equiangular; and will have the angles equal, by which the homologous sides are subtended.

With either leg, DE, from the equal angles of either, A and D, and at either of its extremities, D, construct the angle EDG, equal to the angle A, and at the extremity

E, construct the angle DEG equal to B, and the remaining angle G in the triangle DEG, will be equal to the angle C, in the triangle ABC (by Cor. 2, Prop. 32, B.. 1).

Therefore since ABC is equiangular to the triangle DEG (by Const.), BA shall be to AC, as ED to DG (by Prop. 4, B. 6); but BA is to AC, as ED to DF (by Hypoth.); and therefore ED is to DG, as ED to DF (by Prop. 18, B. 5), therefore DG and DF are equal

(by Prop. 15, B. 5); but the angles EDG and EDF are equal, since either is equal to the angle A, (by Hypoth, and Const.), and the side ED is common to both triangles, therefore the triangle EDF is equiangular to the triangle EDG (by Prop. 4, B. 1), but BAC is equiangular to EDG (by Const.), therefore BAC is also equiangular to EDF; but it is evident that the equal angles are subtended by the homologous sides.

PROPOSITION VII. THEOREM.

If two triangles (ABC, DEF), have one angle of the one equal to one angle of the other (B to E), and the sides about two of the remaining angles proportional (BA to AC, as ED to DF), but the two remaining angles (C and F), be either each less, or each not less than a right angle, the triangles will be equiangular, and shall have the angles equal, about which the sides are proportional.

First, let both angles, C and F, be less than a right angle; and BAC shall be equal to EDF.

For if not, but if it be possible, let either of them, BAC, be greater; and at the point A, with the right line BA, construct the angle BAG, equal to the less EDF.

For, since in the triangles, DEF, ABG, the angles E and B are equal (by Hypoth.), and EDF and BAG. are also equal (by Const.), EFD will be equal to BGA (by Cor. 2, Prop. 32, B. 1); therefore the triangles are equiangular, and BA is to AG as ED to DF (by Prop. 4, B. 6), but BA is to AC, as ED to DF (by Hypoth.), and therefore BA is to AG, as BA to AC (by Prop. 18, B. 5), therefore AG is equal to AC (by Prop. 15, B.] 5); therefore the angle ACG is equal to the angle AGC (by Prop. 5, B, 1), and therefore each is acute (by Cor. Prop. 17, B. 1), and since AGC is acute,

AGB will be obtuse, and EFD equal to AGB, is also obtuse, but it is said to be acute which is absurd.

Therefore the angle BAC is not greater than EDF, and it can be similarly demonstrated that EDF, is not greater than BAC, they are therefore equal; and since the angles ABC and DEF are also equal (by Hypoth.) the triangles shall be equiangular (by Cor. 2, Prop. 32, B. 1.) therefore they have the sides about the equal angles, proportional (by Prop. 4, B. 6). It can be similarly proved if each angle, C and F, be not less than a right angle, that BAC and EDF are equal, and therefore, that the triangles are equiangular, and therefore that they have the sides about the equal angles proportional.

COR. 1. If in two triangles ABC and DEF, an angle B of the one be equal to the angle E of the other, and the sides about another two angles proportional, BA to AC, as ED to DF, and the remaining angle C be right, the triangles will be equiangular; for letting the construction of the preceding Proposition remain, it can be demonstrated that the triangle CAG is isosceles, and therefore that the angle AGC is equal to the right angle C, which is absurd.

COR. 2.-If the sides opposite to any of the equal angles be equal, the triangles themselves will be equal, as appears from Prop. 26, B. 1.

PROPOSITION VIII. THEOREM.

If in a right angled triangle (ABC), a perpendicular (BF) be drawn from the right angle to the opposite side; it divides the triangle into parts both similar to the whole triangle and to each other.

For in the triangles ABF and ABC, the angle AFB is equal to the angle ABC (by Hypoth.), and the angle A is common to both, therefore the remaining angle ABF is equal to the remaining angle C (by Cor. 2, Prop. 32, B. 1), and therefore

F C

the triangles are equiangular: therefore the sides about the equal angles are proportional (by Prop. 4, B. 6), and the triangles are similar (by Def. 1, B. 6).

It can be similarly demonstrated that the triangle FBC is similar to the triangle ABC. Therefore, because the angle ABF is equal to the angle C, and the angles AFB and BFC are also equal, the remaining angle A shall be equal to the remaining angle FBC, therefore the triangles ABF and FBC are equiangular; and therefore the sides about the equal angles are proportional (by Prop. 4, B. 6), therefore the triangles are similar.

COR.-Hence it appears that the perpendicular BF is a mean proportional between the segments AF and FC, of the side on which it falls; and that the remaining sides, AB and BC, are mean proportionals between the conterminous segments AF and FC, and the whole side AC.

PROPOSITION IX. PROBLEM.

From a given right line (AB) to cut off a required part.

From either extremity, A, of the given line, draw AD, making any angle with D AB, and take any point in it C; make AD the same multiple of AC, that AB is of the part to be cut off, and draw BD, draw through C a right line, CI, parallel to BD, AI will be the part required.

For AI is to AB, as AC to AD (by Cor. 1, Prop. 4, B. 6), and therefore because AC is a submultiple of AD (by Const.), AI will be an equi-submultiple of AB (by Prop. 13, B. 5).

PROPOSITION X. PROBLEM.

To divide a given right line (AB) similarly to a given divided line (FG).

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Euclid's Elements of geometry, the first three books (the fourth, fifth, and ...