PROPOSITION V. THEOREM.

If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those produced, proportionally; and conversely, if the sides or the sides produced, be cut proportionally, the cutting line will be parallel to the third side of the triangle.

In the triangle ABC let DE be drawn parallel to BC, then AD: DB:: AE: EC.

For join BE, CD. Then the triangles BDE, DEC are equivalent, for their bases are the same, and being between parallels their altitudes are equal; therefore ADE: BDE:: ADE : DEC, but (Prop. I.), ADE: BDE::AD: DB, and ADE: DEC :: AE : EC; hence (Prop. II. B. V.) AD: DB :: AE: EC.

AA

Conversely. Let now DE cut the sides AB, AC or their production, so that AD: DB:: AE: EC, then DE will be parallel to BC.

For the same construction remaining, AD: DB :: AED : DEB, and AE: EC: AED: DEC, hence AED: DEB:: AED: DEC; consequently (Prop. IX. Cor. 2. B. V.), the triangles DEB, DEC are equivalent, and having the same base DE, their altitudes are equal, that is, they are between the same parallels.

Cor. AD+DB : AD :: AE+EC: AE, that is, AB : AD :: AC: AE, also AB: BD :: AC: CE.

PROPOSITION VI. THEOREM.

If through two sides of a triangle two lines be drawn parallel to the third side, the portions intercepted will be to each other as the sides themselves.

Let the two sides AB, AC of the triangle ABC be divided by the lines DF, EG parallel to BC; then AB AC :: DE: FG.

:

For by last proposition AD: AF:: DE: FG, and by the corollary AD : AF:: AB: AC; consequently, AB : AC DE: FG.

D

E

B

A

Cor. Hence if any number of parallels be drawn, the sides will be cut proportionally, the opposite intercepted portions being to each other as the sides themselves.

Scholium.

The converse of this proposition does not follow, that is, it is not true, that if the portions intercepted by two lines cutting two sides of a triangle are to each other as those sides, the cutting lines will be parallel to the third side of the triangle; for any two lines drawn from D, E to intercept on the other side a portion equal to FG, will be drawn as this converse directs, although only a single pair can be parallel to the side BC.

PROPOSITION VII. THEOREM.

The line which bisects any angle of a triangle divides the opposite side into portions, which are to each other as the adjacent sides.

Let AD bisect the angle A of the triangle ABC, then BD: DC:: AB: AC.

Draw CE parallel to DA, meeting BA produced in E.

Then (Prop. V.) BD: DC :: BA : AE. Now because AD, EC are parallel, the angle E is equal to the angle BAD, and the angle ACE to the angle CAD, which is equal, by hypothesis, to BAD; it appears then that the angles E and ACE are each equal to BAD; therefore AE is equal to AC; hence, putting AC for AE in the above proportion, we have BD: DC :: BA: AC.

B

E

PROPOSITION VIII. THEOREM. (Converse of Prop. VII.) If a line from the vertex of any angle of a triangle divide the side opposite into portions, which are to each other as the sides adjacent, the line so drawn bisects the angle.

In the triangle ABC (preceding diagram) let the line AD divide BC, so that BD: DC :: BA AC, then is the angle BAD equal to the angle CAD.

:

Draw CE parallel to DA, meeting BA, produced in E. Then (Prop. V.) BD : DC :: BA: AE; but by hypothesis, BD: DC: BA: AC, therefore BA AE: BA : AC; consequently, AE AC, and therefore the angle ACE to the angle AEC; but because of the parallels AD, EC, the angles AČE, AEC are respectively equal to the angles DAC, DAB; these angles are therefore equal, and consequently AD bisects the angle BAC.

PROPOSITION IX. THEOREM.

If two triangles have the angles of the one respectively equal to those of the other, the sides containing the equal angles are proportional.

Let the triangles ABC, DCE have the angles A, B, in the one, respectively equal to the angles D, C, in the other; these triangles will be similar.

For let them be placed so that two homologous sides BC, CE may form one straight line, and produce BA, ED till they meet in F.

Then since the angle B is equal to the angle DCE, the line BF is parallel to the line CD; also since the angle ACB is equal to the angle E, the line AC is parallel to the line FE; therefore CF is a rhomboid, and con- B sequently AF-CD, and FD=AC.

A

F

D

E

Now because AC is parallel to FE, we have (Prop. V.) BC: CE: BA: AF, and because CD is parallel to BF we have BC: CE :: FD: DE, and if in these proportions CD be put for its equal AF, and AC for its equal FD, they become BC CE: BA: CD

whence

BC CE: AC: DE

:

BA: CD :: AC: DE.

Therefore the sides containing the equal angles are proportional.

Cor. 1. Hence (Def. 1.), if the angles of one triangle are respectively equal to those of another, the triangles are similar. Cor. 2. Therefore if a line be drawn parallel to one of the sides of a triangle it will cut off a triangle similar to the whole.

PROPOSITION X.

THEOREM.

(Converse of Prop. IX.)

The angles of one triangle are respectively equal to those of another, if the containing sides are proportional.

In the triangles ABC, DEF, let there be the following proportions among the sides, viz.

AB: BC: DE: EF
AC CB: DF: FE;

then will the angles B, C, A be respectively equal to the angles E, F, D.

For, let EG, FG be drawn, making angles at E and F, with the side EF respectively equal to the angles B and C; then the triangles ABC, GEF have the angles in the one respectively equal to those of the other, and, consequently (Prop. IX.), the containing sides are proportional; so that

but

AB: BC: EG: EF;

CE

F

AB: BC :: DE: EF, by hypothesis: hence

(Prop. IX. Cor. 3. B. V.) EG is to equal to DE.

Again,

but

AC: CB:: FG: FE;

AC: CB :: DF: FE, by hypothesis; therefore, FG is equal to DF. Since, then, the sides of the triangle EGF are respectively equal to those of the triangle DEF, these triangles are equal; but the angles of the triangle EGF are respectively equal to those of the triangle ABC; therefore the angles of the triangle DEF are respectively equal to those of the triangle ABC.

Cor. Hence, triangles whose corresponding sides are proportional, are similar. Indeed the above demonstration establishes the similarity of the triangles, if they have the sides containing two angles in each proportional.

PROPOSITION XI. THEOREM.

Triangles are similar, which have an angle in the one equal to an angle in the other, and the containing sides proportional.

Let the triangles ABC, DEF have the angle B in the one equal to the angle E in the other, while the containing sides form the proportion AB: BC:: DE: EF; the triangles are similar.

For, make BG, BH respectively equal to ED, EF, and join GH.

Then the triangles GBH,

DEF are equal, since two B

sides and the included angle in the one are respectively

CE

H

D

equal to two sides and the included angle in the other; hence, and by hypothesis, AB: BC :: GB: BH, that is, the sides BA, BC of the triangle ABC are cut proportionally by the line GH; GH, therefore, is parallel to AC (Prop. V.): hence (Prop. IX. Cor. 2.) the triangle GBH is similar to the triangle ABC, and the triangle DEF has been shown to be equal to the triangle GBH; therefore the triangle DEF is similar to the triangle ABC.

Scholium.

The above proposition is obviously true of rhomboids, that is, rhomboids are similar which have an angle in the one equal to an angle in the other, and the containing sides proportional; for such rhomboids must be equiangular, and the opposite sides of each being equal, it follows that the sides containing the equal angles are proportional; therefore (Def. 1.) they are similar.

PROPOSITION XII. THEOREM.

Triangles are similar which have an angle in each equal, and the sides containing another angle proportional, provided the third angle in each be of the same character.

In the triangles ABC, DEF let the angles B, E be equal, while the angles A, D are both either right, obtuse, or acute;